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## Solve quadratic equations using the formula

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In this Concept, you'll learn to use the quadratic formula to solve quadratic equations.

### Watch This

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

You may have learned one or more of these three other methods to solve a quadratic equation:

• Graphing to find the zeros;
• Solving using square roots; and
• Completing the square to find the solutions

This Concept will present a fourth way to solve a quadratic equation: using 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, 0=ax2+bx+c\begin{align*}0=ax^2+bx+c\end{align*}, is:

x=b±b24ac2a\begin{align*}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{align*}

#### Example A

Solve x2+10x+9=0\begin{align*}x^2+10x+9=0\end{align*} using the quadratic formula.

Solution:

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

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

xxxxxx=10±(10)24(1)(9)2(1)=10±100362=10±642=10±82=10+82 or x=1082=1 or x=9\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 4x2+x+1=0\begin{align*}-4x^2+x+1=0\end{align*} using the quadratic formula.

Solution:

Quadratic formula:Plug in the values a=4,b=1,c=1:Simplify:Separate the two options:Solve:xxxxx=b±b24ac2a=1±(1)24(4)(1)2(4)=1±1+168=1±178=1+178 and x=1178.39 and x.64\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 8t2+10t+3=0\begin{align*}8t^2+10t+3=0\end{align*} using the quadratic formula.

Solution:

Quadratic formula:Plug in the values a=8,b=10,c=3:Simplify:Separate the two options:Separate the two options:Solve:xxxxxx=b±b24ac2a=10±(10)24(8)(3)2(8)=10±100+9616=10±19616=10+19616 and x=1019616=10+1416 and x=101416=14 and x=32\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 3k2+11k=4\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:

3k2+11k=43k2+11k4=0\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.

Quadratic formula:Plug in the values a=3,b=11,c=4:Simplify:Separate the two options:Separate the two options:Solve:xxxxxx=b±b24ac2a=11±(11)24(3)(4)2(3)=11±121+486=11±1696=11+1696 and x=111696=11+136 and x=11136=13 and x=4\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)

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

1. x2+4x21=0\begin{align*}x^2+4x-21=0\end{align*}
2. x26x=12\begin{align*}x^2-6x=12\end{align*}
3. 3x212x=38\begin{align*}3x^2-\frac{1}{2}x=\frac{3}{8}\end{align*}
4. 2x2+x3=0\begin{align*}2x^2+x-3=0\end{align*}
5. x27x+12=0\begin{align*}-x^2-7x+12=0\end{align*}
6. 3x2+5x=0\begin{align*}-3x^2+5x=0\end{align*}
7. 4x2=0\begin{align*}4x^2=0\end{align*}
8. x2+2x+6=0\begin{align*}x^2+2x+6=0\end{align*}

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