Solve Quadratic Equations Using the Quadratic Formula

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 of solutions of a quadratic equation, and identify the most appropriate method to use to solve a quadratic equation.

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 this exercise set, you may have been wondering “isn’t there an easier way to do this?” The answer is “yes.” 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 in general to solve a quadratic equation for $x$ . It may be helpful to look at one of the examples at the end of the last section where we solved an equation of the form $ax^2+bx+c=0$ as you read through the algebraic steps below, so you see them with numbers as well as “in general.”

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

\begin{array}{lrcl} \text{Start with standard form.} & ax^2+bx+c &=& 0,\ a\ne0 \\[4pt] \text{Isolate the variable terms on one side.} & ax^2+bx &=& -c \\[4pt] \text{Make leading coefficient 1, by dividing by }a. & \tfrac{ax^2}{a}+\tfrac ba x &=& -\tfrac ca \\[10pt] \text{Simplify.} & x^2+\tfrac ba x &=& -\tfrac ca \\[10pt] \text{Find }(\tfrac12\cdot\tfrac ba)^2\text{ and add it to both sides.} & x^2+\tfrac ba x+\tfrac{b^2}{4a^2} &=& -\tfrac ca+\tfrac{b^2}{4a^2} \\[10pt] \text{The left side is a perfect square, factor it.} & (x+\tfrac{b}{2a})^2 &=& -\tfrac ca+\tfrac{b^2}{4a^2} \\[10pt] \text{Write equivalent fractions with the common denominator.} & (x+\tfrac{b}{2a})^2 &=& \tfrac{b^2}{4a^2}-\tfrac{c\cdot4a}{a\cdot4a} \\[10pt] \text{Simplify.} & (x+\tfrac{b}{2a})^2 &=& \tfrac{b^2}{4a^2}-\tfrac{4ac}{4a^2} \\[10pt] \text{Combine to one fraction.} & (x+\tfrac{b}{2a})^2 &=& \tfrac{b^2-4ac}{4a^2} \\[10pt] \text{Use the square root property.} & x+\tfrac{b}{2a} &=& \pm\sqrt{\tfrac{b^2-4ac}{4a^2}} \\[10pt] \text{Simplify.} & x+\tfrac{b}{2a} &=& \pm\tfrac{\sqrt{b^2-4ac}}{2a} \\[10pt] \text{Add }-\tfrac{b}{2a}\text{ to both sides.} & x &=& -\tfrac{b}{2a}\pm\tfrac{\sqrt{b^2-4ac}}{2a} \\[10pt] \text{Combine the terms on the right side.} & x &=& \tfrac{-b\pm\sqrt{b^2-4ac}}{2a} \end{array}

This last equation is the Quadratic Formula.

Quadratic Formula. The solutions to a quadratic equation of the form $ax^2+bx+c=0$ , $a\ne0$ are given by the formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

To use the Quadratic Formula we substitute the values of $a$ , $b$ , and $c$ into the expression on the right side of the formula. Then, we do all the math to simplify the expression. The result gives the solution(s) to the quadratic equation.

Example 10.28. How to Solve a Quadratic Equation Using the Quadratic Formula. Solve $2x^2+9x-5=0$ by using the Quadratic Formula.

\begin{array}{lrcl} \text{Write in standard form and identify }a,b,c. & 2x^2+9x-5 &=& 0 \\[4pt] &&& a=2,\ b=9,\ c=-5 \\[4pt] \text{Substitute in the Quadratic Formula.} & x &=& \tfrac{-9\pm\sqrt{9^2-4\cdot2\cdot(-5)}}{2\cdot2} \\[10pt] \text{Simplify the fraction, and solve for }x. & x &=& \tfrac{-9\pm\sqrt{121}}4 \\[10pt] & x &=& \tfrac{-9\pm11}{4} \\[10pt] &&& x=\tfrac12,\ x=-5 \end{array}

Check the solutions: $2(\tfrac12)^2+9(\tfrac12)-5=0\ \checkmark$ and $2(-5)^2+9(-5)-5=0\ \checkmark$ .

Solve a quadratic equation using the Quadratic Formula.

Write the Quadratic Formula in standard form. Identify the $a$ , $b$ , and $c$ values.
Write the Quadratic Formula. Then substitute in the values of $a$ , $b$ , and $c$ .
Simplify.
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 “ $x=$ .”

Solve $3y^2-5y+2=0$ by using the Quadratic Formula. Enter both solutions separated by commas, least to greatest.

Example 10.29. Solve $x^2-6x+5=0$ by using the Quadratic Formula.

\begin{array}{lrcl} \text{This equation is in standard form.} & x^2-6x+5 &=& 0 \\[4pt] \text{Identify the }a,b,c\text{ values.} &&& a=1,\ b=-6,\ c=5 \\[4pt] \text{Write the Quadratic Formula.} & x &=& \tfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\[10pt] \text{Substitute in the values.} & x &=& \tfrac{-(-6)\pm\sqrt{(-6)^2-4\cdot1\cdot5}}{2\cdot1} \\[10pt] \text{Simplify.} & x &=& \tfrac{6\pm\sqrt{16}}2 \\[10pt] & x &=& \tfrac{6\pm4}{2} \\[10pt] \text{Rewrite to show two solutions.} & x &=& \tfrac{6+4}{2},\ x=\tfrac{6-4}{2} \\[10pt] \text{Simplify.} &&& x=5,\ x=1 \end{array}

Check: $5^2-6\cdot5+5=0\ \checkmark$ and $1^2-6\cdot1+5=0\ \checkmark$ .

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 10.30. Solve $4y^2-5y-3=0$ by using the Quadratic Formula.

We can use the Quadratic Formula to solve for the variable in a quadratic equation, whether or not it is named $x$ .

\begin{array}{lrcl} \text{This equation is in standard form.} & 4y^2-5y-3 &=& 0 \\[4pt] \text{Identify the }a,b,c\text{ values.} &&& a=4,\ b=-5,\ c=-3 \\[4pt] \text{Write the Quadratic Formula.} & y &=& \tfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\[10pt] \text{Substitute in the values.} & y &=& \tfrac{-(-5)\pm\sqrt{(-5)^2-4\cdot4\cdot(-3)}}{2\cdot4} \\[10pt] \text{Simplify.} & y &=& \tfrac{5\pm\sqrt{73}}8 \\[10pt] \text{Rewrite to show two solutions.} & y &=& \tfrac{5+\sqrt{73}}8,\ y=\tfrac{5-\sqrt{73}}8 \end{array}

Check. We leave the check to you.

Solve $2p^2+8p+5=0$ by using the Quadratic Formula. Enter both solutions separated by commas, least to greatest.

Example 10.31. Solve $2x^2+10x+11=0$ by using the Quadratic Formula.

\begin{array}{lrcl} \text{Identify the values.} & 2x^2+10x+11 &=& 0,\quad a=2,\ b=10,\ c=11 \\[4pt] \text{Substitute in the formula.} & x &=& \tfrac{-(10)\pm\sqrt{(10)^2-4\cdot2\cdot11}}{2\cdot2} \\[10pt] \text{Simplify.} & x &=& \tfrac{-10\pm\sqrt{12}}4 \\[10pt] \text{Simplify the radical.} & x &=& \tfrac{-10\pm2\sqrt3}{4} \\[10pt] \text{Factor out the common factor.} & x &=& \tfrac{2(-5\pm\sqrt3)}4 \\[10pt] \text{Remove common factors.} & x &=& \tfrac{-5\pm\sqrt3}{2} \\[10pt] \text{Rewrite to show two solutions.} & x &=& \tfrac{-5+\sqrt3}{2},\ x=\tfrac{-5-\sqrt3}{2} \end{array}

Check. We leave the check to you.

We cannot take the square root of a negative number. So, when we substitute $a$ , $b$ , and $c$ into the Quadratic Formula, if the quantity inside the radical is negative, the quadratic equation has no real solution. We will see this in the next example.

Example 10.32. Solve $3p^2+2p+9=0$ by using the Quadratic Formula.

\begin{array}{lrcl} \text{Identify the values.} & 3p^2+2p+9 &=& 0,\quad a=3,\ b=2,\ c=9 \\[4pt] \text{Write the Quadratic Formula.} & p &=& \tfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\[10pt] \text{Substitute in the values.} & p &=& \tfrac{-(2)\pm\sqrt{(2)^2-4\cdot3\cdot9}}{2\cdot3} \\[10pt] \text{Simplify.} & p &=& \tfrac{-2\pm\sqrt{4-108}}6 \\[10pt] \text{Simplify the radical.} & p &=& \tfrac{-2\pm\sqrt{-104}}6 \\[10pt] \text{We cannot take the square root of a negative number.} &&& \text{There is no real solution.} \end{array}

Solve $4a^2-3a+8=0$ by using the Quadratic Formula.

The quadratic equations we have solved so far in this section were all written in standard form, $ax^2+bx+c=0$ . Sometimes, we will need to do some algebra to get the equation into standard form before we can use the Quadratic Formula.

Example 10.33. Solve $x(x+6)+4=0$ by using the Quadratic Formula.

\begin{array}{lrcl} & x(x+6)+4 &=& 0 \\[4pt] \text{Distribute to get standard form.} & x^2+6x+4 &=& 0 \\[4pt] \text{Identify the values.} &&& a=1,\ b=6,\ c=4 \\[4pt] \text{Write the Quadratic Formula.} & x &=& \tfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\[10pt] \text{Substitute in the values.} & x &=& \tfrac{-(6)\pm\sqrt{(6)^2-4\cdot1\cdot4}}{2\cdot1} \\[10pt] \text{Simplify.} & x &=& \tfrac{-6\pm\sqrt{20}}2 \\[10pt] \text{Simplify the radical.} & x &=& \tfrac{-6\pm2\sqrt5}{2} \\[10pt] \text{Factor out the common factor.} & x &=& \tfrac{2(-3\pm\sqrt5)}2 \\[10pt] \text{Remove common factors.} & x &=& -3\pm\sqrt5 \\[4pt] \text{Rewrite to show two solutions.} & x &=& -3+\sqrt5,\ x=-3-\sqrt5 \end{array}

Check. We leave the check to you.

Solve $x(x+2)-5=0$ by using the Quadratic Formula. Enter both solutions separated by commas, least to greatest.

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 10.34. Solve $\tfrac12u^2+\tfrac23u=\tfrac13$ by using the Quadratic Formula.

\begin{array}{lrcl} & \tfrac12u^2+\tfrac23u &=& \tfrac13 \\[10pt] \text{Multiply both sides by the LCD, 6.} & 6(\tfrac12u^2+\tfrac23u) &=& 6(\tfrac13) \\[10pt] \text{Multiply.} & 3u^2+4u &=& 2 \\[4pt] \text{Subtract 2 to get standard form.} & 3u^2+4u-2 &=& 0 \\[4pt] \text{Identify the values.} &&& a=3,\ b=4,\ c=-2 \\[4pt] \text{Write the Quadratic Formula.} & u &=& \tfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\[10pt] \text{Substitute in the values.} & u &=& \tfrac{-(4)\pm\sqrt{(4)^2-4\cdot3\cdot(-2)}}{2\cdot3} \\[10pt] \text{Simplify.} & u &=& \tfrac{-4\pm\sqrt{40}}6 \\[10pt] \text{Simplify the radical.} & u &=& \tfrac{-4\pm2\sqrt{10}}6 \\[10pt] \text{Factor out the common factor.} & u &=& \tfrac{2(-2\pm\sqrt{10})}6 \\[10pt] \text{Remove common factors.} & u &=& \tfrac{-2\pm\sqrt{10}}3 \\[10pt] \text{Rewrite to show two solutions.} & u &=& \tfrac{-2+\sqrt{10}}3,\ u=\tfrac{-2-\sqrt{10}}3 \end{array}

Check. We leave the check to you.

Solve $\tfrac14c^2-\tfrac13c=\tfrac1{12}$ by using the Quadratic Formula. Enter both solutions separated by commas, least to greatest.

Think about the equation $(x-3)^2=0$ . We know from the Zero Products Principle 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 with a perfect square also gives just one solution.

Example 10.35. Solve $4x^2-20x=-25$ by using the Quadratic Formula.

\begin{array}{lrcl} & 4x^2-20x &=& -25 \\[4pt] \text{Add 25 to get standard form.} & 4x^2-20x+25 &=& 0 \\[4pt] \text{Identify the values.} &&& a=4,\ b=-20,\ c=25 \\[4pt] \text{Write the Quadratic Formula.} & x &=& \tfrac{-b\pm\sqrt{b^2-4ac}}{2a} \\[10pt] \text{Substitute in the values.} & x &=& \tfrac{-(-20)\pm\sqrt{(-20)^2-4\cdot4\cdot25}}{2\cdot4} \\[10pt] \text{Simplify.} & x &=& \tfrac{20\pm\sqrt0}{8} \\[10pt] \text{Simplify the radical.} & x &=& \tfrac{20}{8} \\[10pt] \text{Simplify the fraction.} & x &=& \tfrac52 \end{array}

Check. We leave the check to you. Did you recognize that $4x^2-20x+25$ is a perfect square?

Solve $r^2+10r+25=0$ by using the Quadratic Formula.

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

When we solved the quadratic equations in the previous examples, sometimes we got two solutions, sometimes one solution, sometimes no real solutions. Is there a way to predict the number of solutions to a quadratic equation without actually solving the equation?

Yes, the quantity inside the radical of the Quadratic Formula makes it easy for us to determine the number of solutions. This quantity is called the discriminant.

Discriminant. In the Quadratic Formula

x=\tfrac{-b\pm\sqrt{b^2-4ac}}{2a}

, the quantity

b^2-4ac

is called the discriminant.

Let’s look at the discriminant of the equations in Example 10.28, Example 10.32, and Example 10.35, and the number of solutions to those quadratic equations.

Quadratic equation (in standard form)	Discriminant $b^2-4ac$	Sign	Number of real solutions
$2x^2+9x-5=0$	$9^2-4\cdot2(-5)=121$	$+$	2
$4x^2-20x+25=0$	$(-20)^2-4\cdot4\cdot25=0$	$0$	1
$3p^2+2p+9=0$	$2^2-4\cdot3\cdot9=-104$	$-$	0

When the discriminant is positive, the quadratic equation has two solutions. When the discriminant is zero, the quadratic equation has one solution. When the discriminant is negative, the quadratic equation has no real solutions.

Use the discriminant, $b^2-4ac$ , to determine the number of solutions of a Quadratic Equation. For $ax^2+bx+c=0$ , $a\ne0$ :

if $b^2-4ac>0$ , the equation has two solutions.
if $b^2-4ac=0$ , the equation has one solution.
if $b^2-4ac<0$ , the equation has no real solutions.

Example 10.36. Determine the number of solutions to each quadratic equation: (a) $2v^2-3v+6=0$ (b) $3x^2+7x-9=0$ (c) $5n^2+n+4=0$ (d) $9y^2-6y+1=0$ .

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

(a) For $2v^2-3v+6=0$ , $a=2$ , $b=-3$ , and $c=6$ .

b^2-4ac=(-3)^2-4\cdot2\cdot6=9-48=-39

Because the discriminant is negative, there are no real solutions to the equation.

(b) For $3x^2+7x-9=0$ , $a=3$ , $b=7$ , and $c=-9$ .

b^2-4ac=7^2-4\cdot3\cdot(-9)=49+108=157

Because the discriminant is positive, there are two solutions to the equation.

(c) For $5n^2+n+4=0$ , $a=5$ , $b=1$ , and $c=4$ .

b^2-4ac=1^2-4\cdot5\cdot4=1-80=-79

Because the discriminant is negative, there are no real solutions to the equation.

(d) For $9y^2-6y+1=0$ , $a=9$ , $b=-6$ , and $c=1$ .

b^2-4ac=(-6)^2-4\cdot9\cdot1=36-36=0

Because the discriminant is 0, there is one solution to the equation.

Determine the number of solutions to $8m^2-3m+6=0$ .

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

We have used four methods to solve quadratic equations:

Factoring
Square Root Property
Completing the Square
Quadratic Formula

You can solve any quadratic equation by using the Quadratic Formula, but that is not always the easiest method to use.

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 quadratic equation can be solved by using the Quadratic Formula.

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 this chapter 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.

Example 10.37. Identify the most appropriate method to use to solve each quadratic equation: (a) $5z^2=17$ (b) $4x^2-12x+9=0$ (c) $8u^2+6u=11$ .

(a) $5z^2=17$

Since the equation is in the $ax^2=k$ form, the most appropriate method is to use the Square Root Property.

(b) $4x^2-12x+9=0$

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

(c) $8u^2+6u=11$

Put the equation in standard form: $8u^2+6u-11=0$ .

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

Identify the most appropriate method to use to solve $x^2+6x+8=0$ .

Key terms

Quadratic Formula — the solutions to a quadratic equation of the form $ax^2+bx+c=0$ , $a\ne0$ , are given by $x=\tfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ . discriminant — the quantity $b^2-4ac$ in the Quadratic Formula; its sign predicts the number of real solutions.

This page is adapted from Elementary Algebra 2e, Section 10.3 by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: reformatted the source for accessible web presentation and converted selected Try It problems into interactive exercises; the source exercise set and media links are omitted.

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