# 10.7: Quadratic Formula

**Basic**Created by: CK-12

**Practice**Quadratic Formula

What if you needed to solve the quadratic equation \begin{align*}x^2+x-5=0\end{align*} in order to determine the width of a rectangular map? You graphed the function \begin{align*}f(x)=x^2+x-5\end{align*}, but from the graph you can only tell the approximate width, and you want a more precise answer. In this Concept, you'll learn to use the quadratic formula to solve quadratic equations like the one representing this situation so that you can get exact solutions to the equations.

### Watch This

**Multimedia Link** For more examples of solving quadratic equations using the quadratic formula, see Khan Academy Equation Part 2 (9:14).

**Figure** 2 provides more examples of solving equations using the quadratic equation. This video is not necessarily different from the examples above, but it does help reinforce the procedure of using the quadratic formula to solve equations.

### Guidance

Previous Concepts have presented three methods to solve a quadratic equation:

- By graphing to find the zeros;
- By solving using square roots; and
- By using completing the square to find the solutions

This Concept will present a fourth way to solve a quadratic equation: using the quadratic formula.

**History of the Quadratic Formula**

As early as 1200 BC, people were interested in solving quadratic equations. The Babylonians solved simultaneous equations involving quadratics. In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit formula to solve a quadratic equation. The quadratic formula was written as it is today by the Arabic mathematician Al-Khwarizmi. It is his name upon which the word “Algebra” is based.

The solution to any quadratic equation in standard form, \begin{align*}0=ax^2+bx+c\end{align*}, is:

\begin{align*}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{align*}

#### Example A

*Solve \begin{align*}x^2+10x+9=0\end{align*} using the quadratic formula.*

**Solution:**

We know from the last Concept the answers are \begin{align*}x=-1\end{align*} *or* \begin{align*}x=-9\end{align*}.

By applying the quadratic formula and \begin{align*}a=1, b=10\end{align*}, and \begin{align*}c=9\end{align*}, we get:

\begin{align*}x &= \frac{-10 \pm \sqrt{(10)^2-4(1)(9)}}{2(1)}\\ x &= \frac{-10 \pm \sqrt{100-36}}{2}\\ x &= \frac{-10 \pm \sqrt{64}}{2}\\ x &= \frac{-10 \pm 8}{2}\\ x &= \frac{-10 + 8}{2} \ or \ x=\frac{-10-8}{2}\\ x &= -1 \ or \ x=-9\end{align*}

#### Example B

*Solve* \begin{align*}-4x^2+x+1=0\end{align*} *using the quadratic formula.*

**Solution:**

\begin{align*}\text{Quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \text{Plug in the values} \ a=-4, b=1, c=1: && x& =\frac{-1 \pm \sqrt{(1)^2-4(-4)(1)}}{2(-4)}\\ \text{Simplify:} && x & =\frac{-1 \pm \sqrt{1+16}}{-8}=\frac{-1 \pm \sqrt{17}}{-8}\\ \text{Separate the two options:} && x&=\frac{-1+\sqrt{17}}{-8} \ \text{and} \ x=\frac{-1-\sqrt{17}}{-8}\\ \text{Solve:} && x & \approx -.39 \ \text{and} \ x \approx .64\end{align*}

#### Example C

*Solve* \begin{align*}8t^2+10t+3=0\end{align*} *using the quadratic formula.*

**Solution:**

\begin{align*}\text{Quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \text{Plug in the values} \ a=8, b=10, c=3: && x& =\frac{-10 \pm \sqrt{(10)^2-4(8)(3)}}{2(8)}\\ \text{Simplify:} && x & =\frac{-10 \pm \sqrt{100+96}}{16}=\frac{-10 \pm \sqrt{196}}{16}\\ \text{Separate the two options:} && x&=\frac{-10+\sqrt{196}}{16} \ \text{and} \ x=\frac{-10-\sqrt{196}}{16}\\ \text{Separate the two options:} && x&=\frac{-10+14}{16} \ \text{and} \ x=\frac{-10-14}{16}\\ \text{Solve:} && x & =\frac{1}{4} \ \text{and} \ x =-\frac{3}{2}\end{align*}

### Guided Practice

*Solve* \begin{align*}3k^2+11k=4\end{align*} *using the quadratic formula.*

**Solution:**

First, we must make it so one side is equal to zero:

\begin{align*}3k^2+11k=4 \Rightarrow 3k^2+11k-4=0\end{align*}

Now it is in the correct form for using the quadratic formula.

\begin{align*}\text{Quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \text{Plug in the values} \ a=3, b=11, c=-4: && x& =\frac{-11 \pm \sqrt{(11)^2-4(3)(-4)}}{2(3)}\\ \text{Simplify:} && x & =\frac{-11 \pm \sqrt{121+48}}{6}=\frac{-11 \pm \sqrt{169}}{6}\\ \text{Separate the two options:} && x&=\frac{-11+\sqrt{169}}{6} \ \text{and} \ x=\frac{-11-\sqrt{169}}{6}\\ \text{Separate the two options:} && x&=\frac{-11+13}{6} \ \text{and} \ x=\frac{-11-13}{6}\\ \text{Solve:} && x & =\frac{1}{3} \ \text{and} \ x =-4\end{align*}

### Practice

The following video will guide you through a proof of the quadratic formula. CK-12 Basic Algebra: Proof of Quadratic Formula (7:44)

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Using the Quadratic Formula (16:32)

- What is the quadratic formula? When is the most appropriate situation to use this formula?
- When was the first known solution of a quadratic equation recorded?

Solve the following quadratic equations using the quadratic formula.

- \begin{align*}x^2+4x-21=0\end{align*}
- \begin{align*}x^2-6x=12\end{align*}
- \begin{align*}3x^2-\frac{1}{2}x=\frac{3}{8}\end{align*}
- \begin{align*}2x^2+x-3=0\end{align*}
- \begin{align*}-x^2-7x+12=0\end{align*}
- \begin{align*}-3x^2+5x=0\end{align*}
- \begin{align*}4x^2=0\end{align*}
- \begin{align*}x^2+2x+6=0\end{align*}

**Mixed Review**

- The theater has three types of seating: balcony, box, and floor. There are four times as many floor seats as balcony. There are 200 more box seats than balcony seats. The theater has a total of 1,100 seats. Determine the number of balcony, box, and floor seats in the theater.
- Write an equation in slope-intercept form containing (10, 65) and (5, 30).
- 120% of what number is 60?
- Name the set(s) of numbers to which \begin{align*}\sqrt{16}\end{align*} belongs.
- Divide: \begin{align*}6 \frac{1}{7} \div - 2 \frac{3}{4}\end{align*}.
- The set is the number of books in a library. Which of the following is the most appropriate domain for this set: all real numbers; positive real numbers; integers; or whole numbers? Explain your reasoning.

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Term | Definition |
---|---|

Quadratic Formula |
The quadratic formula states that for any quadratic equation in the form , . |

Binomial |
A binomial is an expression with two terms. The prefix 'bi' means 'two'. |

Completing the Square |
Completing the square is a common method for rewriting quadratics. It refers to making a perfect square trinomial by adding the square of 1/2 of the coefficient of the term. |

Roots |
The roots of a function are the values of x that make y equal to zero. |

Square Root |
The square root of a term is a value that must be multiplied by itself to equal the specified term. The square root of 9 is 3, since 3 * 3 = 9. |

Vertex |
The vertex of a parabola is the highest or lowest point on the graph of a parabola. The vertex is the maximum point of a parabola that opens downward and the minimum point of a parabola that opens upward. |

### Image Attributions

Here you'll learn how to solve quadratic equations by using the quadratic formula.