how to solve 2 quadratic equations

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How to Solve a Quadratic Equation: A Step-by-Step Guide

Last Updated: May 3, 2024 Fact Checked

Factoring the Equation

Using the quadratic formula, completing the square, practice problems and answers, expert q&a.

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,413,287 times.

A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. [1] X Research source There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.

Quadradic Formula for Solving Equations

• Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4. [3] X Research source

• 3x = -1 ..... by subtracting
• 3x/3 = -1/3 ..... by dividing
• x = -1/3 ..... simplified
• x = 4 ..... by subtracting
• x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.

• So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.

• 4x 2 - 5x - 13 = x 2 -5
• 4x 2 - x 2 - 5x - 13 +5 = 0
• 3x 2 - 5x - 8 = 0

• {-b +/-√ (b 2 - 4ac)}/2
• {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3)

• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
• {5 +/-√(25 + 96)}/6
• {5 +/-√(121)}/6

• (5 + 11)/6 = 16/6
• (5-11)/6 = -6/6

• x = (-1, 8/3)

• 2x 2 - 9 = 12x =
• In this equation, the a term is 2, the b term is -12, and the c term is -9.

• 2x 2 - 12x - 9 = 0
• 2x 2 - 12x = 9

• 2x 2 /2 - 12x/2 = 9/2 =
• x 2 - 6x = 9/2

• -6/2 = -3 =
• (-3) 2 = 9 =
• x 2 - 6x + 9 = 9/2 + 9

• x = 3 + 3(√6)/2
• x = 3 - 3(√6)/2)

• If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example: [14] X Research source Thanks Helpful 0 Not Helpful 0
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 3 Not Helpful 0
• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. Thanks Helpful 1 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
• ↑ http://www.mathsisfun.com/algebra/factoring-quadratics.html
• ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
• ↑ https://www.cuemath.com/algebra/quadratic-equations/
• ↑ https://www.purplemath.com/modules/solvquad4.htm
• ↑ http://www.purplemath.com/modules/quadform.htm
• ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
• ↑ http://www.mathsisfun.com/algebra/completing-square.html
• ↑ http://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm

About This Article

To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down! Did this summary help you? Yes No

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How to Solve Quadratic Equations

How to Solve Quadratic Equations? There are basically three methods to solve quadratic equations. They are:

• Using Quadratic formula
• Factoring the quadratic equation
• Completing the square

A quadratic equation is an equation that has the highest degree equal to two. The standard form of the quadratic equation is ax 2 + bx + c = 0, where a, b, c are constants and a ≠ b ≠ 0. Here, x is an unknown variable for which we need to find the solution. Let us learn here how to solve quadratic equations.

Solving Quadratic Equations – Using Quadratic Formula

The quadratic formula is used to find solutions of quadratic equations. If ax 2 + bx + c = 0, then solution can be evaluated using the formula given below;

Thus, the formula will result in two solutions here.

Here, b 2 – 4ac is the discriminant (D).

D = b 2 – 4ac

• If D = 0, the two roots of quadratic equation are real and equal
• If D > 0, the roots are real and unequal
• If D < 0, the roots are not real, i.e. imaginary

Example: Solve x 2 – 5x + 6 = 0.

Solution: Given,

x 2 – 5x + 6 = 0

a = 1, b = -5, c = 6

By the quadratic formula, we know;

b 2 – 4ac = (-5) 2 – 4 × 1 × 6 = 25 – 24 = 1 > 0

Thus, the roots are real.

x = [-b ± √(b 2 – 4ac)]/ 2a

= [-(-5) ± √1]/ 2(1)

= [5 ± 1]/ 2

i.e. x = (5 + 1)/2 and x = (5 – 1)/2

x = 6/2, x = 4/2

Therefore, the solution of x 2 – 5x + 6 = 0 is 3 or 2.

Solving Quadratic Equations – By Factorisation

We can write the quadratic equation as a product of factors having degree less than or equal to two. This method of solving quadratic equations is called factoring the quadratic equation.

Let us learn by an example.

Example: Solve 6m 2 – 7m + 2 = 0 by factoring method.

Solution: 6m 2 – 4m – 3m + 2 = 0

⟹ 2m(3m – 2) – 1(3m – 2) = 0

⟹ (3m – 2) (2m – 1) = 0

⟹ 3m – 2 = 0 or 2m – 1 = 0

⟹ 3m = 2 or 2m = 1

Therefore, the solutions of the given equation are:

m = ⅔ or m = ½

Solving Quadratic Equation – Completing Square

To solve the quadratic equation using completing the square method, follow the below given steps.

• First make sure the equation is in the standard form: ax 2 + bx + c = 0
• Now, divide the whole equation by a, such that the coefficient of x 2 is 1.
• Write the equation with a constant term on the Right side of equation
• Add the square of half of coefficient of x on both sides and complete the square
• Write the left side equation as a square term and solve

Let us understand with the help of an example.

Example: Solve 4x 2 + x = 3 by completing the square method.

4x 2 + x = 3

Divide the whole equation by 4.

x 2 + x/4 = ¾

Coefficient of x is ¼

Half of ¼ = ⅛

Square of ⅛ = (⅛) 2

Add (⅛) 2 on both sides of the equation.

x 2 + x/4 + (⅛) 2 = ¾ – (⅛) 2

x 2 + x/4 + 1/64 = ¾ + 1/64

(x + ⅛) 2 = (48+1)/64 = 49/64

Taking square root on both sides, we get;

x+1/8 = √(49/64) = ±7/8

x = ⅞ – ⅛ = 6/8 = ⅜

x = – ⅞ – ⅛ = -8/8 = -1

Hence, the solution of quadratic equation is x = ⅜ or x = -1.

Practice Questions

Solve the following quadratic equations.

• 2s 2 + 5s = 3
• 9x 2 – 6x – 2 = 0
• 3x 2 – 11x – 4 = 0
• 2x 2 – 12x – 9 = 0

Frequently Asked Questions on Solving Quadratic Equations

What are the methods to solve quadratic equations, what is the formula to solve quadratic equations.

The quadratic formula is given by: x = [b±√(b²-4ac))]/(2a)

What is the discriminant of the quadratic formula?

The discriminant of quadratic formula is b2 – 4ac. It helps to determine the nature of the roots.

When is the quadratic equation has imaginary roots?

When discriminant of the quadratic equation is less than zero, then the roots are imaginary or non-real.

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11.4: Solve Quadratic Equations Using the Quadratic Formula

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Learning Objectives

By the end of this section, you will be able to:

• Solve quadratic equations using the Quadratic Formula
• Use the discriminant to predict the number and type of solutions of a quadratic equation
• Identify the most appropriate method to use to solve a quadratic equation

Before you get started, take this readiness quiz.

• Evaluate \(b^{2}-4 a b\) when \(a=3\) and \(b=−2\).
• Simplify \(\sqrt{108}\).
• Simplify \(\sqrt{50}\).

Solve Quadratic Equations Using the Quadratic Formula

When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.

We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for \(x\) .

We start with the standard form of a quadratic equation and solve it for \(x\) by completing the square.

The final equation is called the "Quadratic Formula."

Definition \(\PageIndex{1}\): Quadratic Formula

The solutions to a quadratic equation of the form \(a x^{2}+b x+c=0\), where \(a≠0\) are given by the formula:

\[x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \label{quad}\]

To use the Quadratic Formula , we substitute the values of \(a,b\), and \(c\) from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.

Notice the Quadratic Formula (Equation \ref{quad}) is an equation. Make sure you use both sides of the equation.

Example \(\PageIndex{1}\) How to Solve a Quadratic Equation Using the Quadratic Formula

Solve by using the Quadratic Formula: \(2 x^{2}+9 x-5=0\).

Exercise \(\PageIndex{1}\)

Solve by using the Quadratic Formula: \(3 y^{2}-5 y+2=0\).

\(y=1, y=\dfrac{2}{3}\)

Exercise \(\PageIndex{2}\)

Solve by using the Quadratic Formula: \(4 z^{2}+2 z-6=0\).

\(z=1, z=-\dfrac{3}{2}\)

HowTo: Solve a Quadratic Equation Using the Quadratic Formula

• Write the quadratic equation in standard form, \(a x^{2}+b x+c=0\). Identify the values of \(a,b\), and \(c\).
• Write the Quadratic Formula. Then substitute in the values of \(a,b\), and \(c\).
• Check the solutions.

If you say the formula as you write it in each problem, you’ll have it memorized in no time! And remember, the Quadratic Formula is an EQUATION. Be sure you start with “\(x=\)”.

Example \(\PageIndex{2}\)

Solve by using the Quadratic Formula: \(x^{2}-6 x=-5\).

Exercise \(\PageIndex{3}\)

Solve by using the Quadratic Formula: \(a^{2}-2 a=15\).

\(a=-3, a=5\)

Exercise \(\PageIndex{4}\)

Solve by using the Quadratic Formula: \(b^{2}+24=-10 b\).

\(b=-6, b=-4\)

When we solved quadratic equations by using the Square Root Property, we sometimes got answers that had radicals. That can happen, too, when using the Quadratic Formula . If we get a radical as a solution, the final answer must have the radical in its simplified form.

Example \(\PageIndex{3}\)

Solve by using the Quadratic Formula: \(2 x^{2}+10 x+11=0\).

Exercise \(\PageIndex{5}\)

Solve by using the Quadratic Formula: \(3 m^{2}+12 m+7=0\).

\(m=\dfrac{-6+\sqrt{15}}{3}, m=\dfrac{-6-\sqrt{15}}{3}\)

Exercise \(\PageIndex{6}\)

Solve by using the Quadratic Formula: \(5 n^{2}+4 n-4=0\).

\(n=\dfrac{-2+2 \sqrt{6}}{5}, n=\dfrac{-2-2 \sqrt{6}}{5}\)

When we substitute \(a, b\), and \(c\) into the Quadratic Formula and the radicand is negative, the quadratic equation will have imaginary or complex solutions. We will see this in the next example.

Example \(\PageIndex{4}\)

Solve by using the Quadratic Formula: \(3 p^{2}+2 p+9=0\).

Exercise \(\PageIndex{7}\)

Solve by using the Quadratic Formula: \(4 a^{2}-2 a+8=0\).

\(a=\dfrac{1}{4}+\dfrac{\sqrt{31}}{4} i, \quad a=\dfrac{1}{4}-\dfrac{\sqrt{31}}{4} i\)

Exercise \(\PageIndex{8}\)

Solve by using the Quadratic Formula: \(5 b^{2}+2 b+4=0\).

\(b=-\dfrac{1}{5}+\dfrac{\sqrt{19}}{5} i, \quad b=-\dfrac{1}{5}-\dfrac{\sqrt{19}}{5} i\)

Remember, to use the Quadratic Formula, the equation must be written in standard form, \(a x^{2}+b x+c=0\). Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Example \(\PageIndex{5}\)

Solve by using the Quadratic Formula: \(x(x+6)+4=0\).

Our first step is to get the equation in standard form.

Exercise \(\PageIndex{9}\)

Solve by using the Quadratic Formula: \(x(x+2)−5=0\).

\(x=-1+\sqrt{6}, x=-1-\sqrt{6}\)

Exercise \(\PageIndex{10}\)

Solve by using the Quadratic Formula: \(3y(y−2)−3=0\).

\(y=1+\sqrt{2}, y=1-\sqrt{2}\)

When we solved linear equations, if an equation had too many fractions we cleared the fractions by multiplying both sides of the equation by the LCD. This gave us an equivalent equation—without fractions— to solve. We can use the same strategy with quadratic equations.

Example \(\PageIndex{6}\)

Solve by using the Quadratic Formula: \(\dfrac{1}{2} u^{2}+\dfrac{2}{3} u=\dfrac{1}{3}\).

Our first step is to clear the fractions.

Exercise \(\PageIndex{11}\)

Solve by using the Quadratic Formula: \(\dfrac{1}{4} c^{2}-\dfrac{1}{3} c=\dfrac{1}{12}\).

\(c=\dfrac{2+\sqrt{7}}{3}, \quad c=\dfrac{2-\sqrt{7}}{3}\)

Exercise \(\PageIndex{12}\)

Solve by using the Quadratic Formula: \(\dfrac{1}{9} d^{2}-\dfrac{1}{2} d=-\dfrac{1}{3}\).

\(d=\dfrac{9+\sqrt{33}}{4}, d=\dfrac{9-\sqrt{33}}{4}\)

Think about the equation \((x-3)^{2}=0\). We know from the Zero Product Property that this equation has only one solution, \(x=3\).

We will see in the next example how using the Quadratic Formula to solve an equation whose standard form is a perfect square trinomial equal to \(0\) gives just one solution. Notice that once the radicand is simplified it becomes \(0\), which leads to only one solution.

Example \(\PageIndex{7}\)

Solve by using the Quadratic Formula: \(4 x^{2}-20 x=-25\).

Did you recognize that \(4 x^{2}-20 x+25\) is a perfect square trinomial. It is equivalent to \((2 x-5)^{2}\)? If you solve \(4 x^{2}-20 x+25=0\) by factoring and then using the Square Root Property, do you get the same result?

Exercise \(\PageIndex{13}\)

Solve by using the Quadratic Formula: \(r^{2}+10 r+25=0\).

Exercise \(\PageIndex{14}\)

Solve by using the Quadratic Formula: \(25 t^{2}-40 t=-16\).

\(t=\dfrac{4}{5}\)

Use the Discriminant to Predict the Number and Type of Solutions of a Quadratic Equation

When we solved the quadratic equations in the previous examples, sometimes we got two real solutions, one real solution, and sometimes two complex solutions. Is there a way to predict the number and type of solutions to a quadratic equation without actually solving the equation?

Yes, the expression under the radical of the Quadratic Formula makes it easy for us to determine the number and type of solutions. This expression is called the discriminant .

Definition \(\PageIndex{2}\)

Discriminant

Let’s look at the discriminant of the equations in some of the examples and the number and type of solutions to those quadratic equations.

Using the Discriminant \(b^{2}-4ac\), to Determine the Number and Type of Solutions of a Quadratic Equation

For a quadratic equation of the form \(ax^{2}+bx+c=0\), \(a \neq 0\),

• If \(b^{2}-4 a c>0\), the equation has \(2\) real solutions.
• if \(b^{2}-4 a c=0\), the equation has \(1\) real solution.
• if \(b^{2}-4 a c<0\), the equation has \(2\) complex solutions.

Example \(\PageIndex{8}\)

Determine the number of solutions to each quadratic equation.

\(3 x^{2}+7 x-9=0\)

\(5 n^{2}+n+4=0\)

\(9 y^{2}-6 y+1=0\)

To determine the number of solutions of each quadratic equation, we will look at its discriminant.

The equation is in standard form, identify \(a, b\), and \(c\).

\(a=3, \quad b=7, \quad c=-9\)

Write the discriminant.

\(b^{2}-4 a c\)

Substitute in the values of \(a, b\), and \(c\).

\((7)^{2}-4 \cdot 3 \cdot(-9)\)

\(49+108\) \(157\)

Since the discriminant is positive, there are \(2\) real solutions to the equation.

\(a=5, \quad b=1, \quad c=4\)

\((1)^{2}-4 \cdot 5 \cdot 4\)

\(1-80\) \(-79\)

Since the discriminant is negative, there are \(2\) complex solutions to the equation.

\(a=9, \quad b=-6, \quad c=1\)

\((-6)^{2}-4 \cdot 9 \cdot 1\)

\(36-36\) \(0\)

Since the discriminant is \(0\), there is \(1\) real solution to the equation.

Exercise \(\PageIndex{15}\)

Determine the number and type of solutions to each quadratic equation.

• \(8 m^{2}-3 m+6=0\)
• \(5 z^{2}+6 z-2=0\)
• \(9 w^{2}+24 w+16=0\)
• \(2\) complex solutions
• \(2\) real solutions
• \(1\) real solution

Exercise \(\PageIndex{16}\)

• \(b^{2}+7 b-13=0\)
• \(5 a^{2}-6 a+10=0\)
• \(4 r^{2}-20 r+25=0\)

Identify the Most Appropriate Method to Use to Solve a Quadratic Equation

We summarize the four methods that we have used to solve quadratic equations below.

Methods for Solving Quadratic Equations

• Square Root Property
• Completing the Square
• Quadratic Formula

Given that we have four methods to use to solve a quadratic equation, how do you decide which one to use? Factoring is often the quickest method and so we try it first. If the equation is \(ax^{2}=k\) or \(a(x−h)^{2}=k\) we use the Square Root Property. For any other equation, it is probably best to use the Quadratic Formula. Remember, you can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method.

What about the method of Completing the Square? Most people find that method cumbersome and prefer not to use it. We needed to include it in the list of methods because we completed the square in general to derive the Quadratic Formula. You will also use the process of Completing the Square in other areas of algebra.

Identify the Most Appropriate Method to Solve a Quadratic Equation

• Try Factoring first. If the quadratic factors easily, this method is very quick.
• Try the Square Root Property next. If the equation fits the form \(ax^{2}=k\) or \(a(x−h)^{2}=k\), it can easily be solved by using the Square Root Property.
• Use the Quadratic Formula . Any other quadratic equation is best solved by using the Quadratic Formula.

The next example uses this strategy to decide how to solve each quadratic equation.

Example \(\PageIndex{9}\)

Identify the most appropriate method to use to solve each quadratic equation.

• \(5 z^{2}=17\)

\(4 x^{2}-12 x+9=0\)

\(8 u^{2}+6 u=11\)

\(5z^{2}=17\)

Since the equation is in the \(ax^{2}=k\), the most appropriate method is to use the Square Root Property.

We recognize that the left side of the equation is a perfect square trinomial, and so factoring will be the most appropriate method.

Put the equation in standard form.

\(8 u^{2}+6 u-11=0\)

While our first thought may be to try factoring, thinking about all the possibilities for trial and error method leads us to choose the Quadratic Formula as the most appropriate method.

Exercise \(\PageIndex{17}\)

• \(x^{2}+6 x+8=0\)
• \((n-3)^{2}=16\)
• \(5 p^{2}-6 p=9\)

Exercise \(\PageIndex{18}\)

• \(8 a^{2}+3 a-9=0\)
• \(4 b^{2}+4 b+1=0\)
• \(5 c^{2}=125\)
• Factoring or Square Root Property

Access these online resources for additional instruction and practice with using the Quadratic Formula.

• Using the Quadratic Formula
• Solve a Quadratic Equation Using the Quadratic Formula with Complex Solutions
• Discriminant in Quadratic Formula

Key Concepts

\(x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

• Write the quadratic equation in standard form, \(a x^{2}+b x+c=0\). Identify the values of \(a, b, c\).
• Write the Quadratic Formula. Then substitute in the values of \(a, b, c\).
• If \(b^{2}-4 a c=0\), the equation has \(1\) real solution.
• If \(b^{2}-4 a c<0\), the equation has \(2\) complex solutions.
• Try the Square Root Property next. If the equation fits the form \(a x^{2}=k\) or \(a(x-h)^{2}=k\), it can easily be solved by using the Square Root Property.
• Use the Quadratic Formula. Any other quadratic equation is best solved by using the Quadratic Formula.

Quadratic Equation Solver

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Step-By-Step Example

Example (click to try), choose your method, solve by factoring.

Example: 3x^2-2x-1=0

Complete The Square

Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.)

Take the Square Root

Example: 2x^2=18

Quadratic Formula

Example: 4x^2-2x-1=0

About quadratic equations

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Quadratic Equation Solver

We can help you solve an equation of the form " ax 2 + bx + c = 0 " Just enter the values of a, b and c below :

Is it Quadratic?

Only if it can be put in the form ax 2 + bx + c = 0 , and a is not zero .

The name comes from "quad" meaning square, as the variable is squared (in other words x 2 ).

These are all quadratic equations in disguise:

How Does this Work?

The solution(s) to a quadratic equation can be calculated using the Quadratic Formula :

The "±" means we need to do a plus AND a minus, so there are normally TWO solutions !

The blue part ( b 2 - 4ac ) is called the "discriminant", because it can "discriminate" between the possible types of answer:

• when it is positive, we get two real solutions,
• when it is zero we get just ONE solution,
• when it is negative we get complex solutions.

Learn more at Quadratic Equations

Quadratic Formula Calculator

Enter the equation you want to solve using the quadratic formula.

The Quadratic Formula Calculator finds solutions to quadratic equations with real coefficients. For equations with real solutions, you can use the graphing tool to visualize the solutions.

Quadratic Formula : x = − b ± b 2 − 4 a c 2 a

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Solve Using the Quadratic Formula Apply the Quadratic Formula

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Solve Using the Quadratic Formula x 2 + 5 x + 6 = 0 Solve Using the Quadratic Formula x 2 - 9 = 0 Solve Using the Quadratic Formula 5 x 2 - 7 x - 3 = 0 Apply the Quadratic Formula x 2 - 14 x + 49 Apply the Quadratic Formula x 2 - 18 x - 4

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Quadratic Equation Worksheets (pdfs)

Free worksheets with answer keys.

Enjoy these free sheets. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. Plus each one comes with an answer key.

• Solve Quadratic Equations by Factoring
• Solve Quadratic Equations by Completing the Square
• Quadratic Formula Worksheet (real solutions)
• Quadratic Formula Worksheet (complex solutions)
• Quadratic Formula Worksheet (both real and complex solutions)
• Discriminant Worksheet
• Sum and Product of Roots
• Radical Equations Worksheet

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