Precalculus Examples

x^{2} + 7 x + 10 = 0

$x^{2} + 7 x + 10 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values

a = 1

$a = 1$ ,

b = 7

$b = 7$ , and

c = 10

$c = 10$ into the quadratic formula and solve for

x

$x$ .

\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}

$\frac{- 7 \pm \sqrt{7^{2} - 4 \cdot (1 \cdot 10)}}{2 \cdot 1}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

7

$7$ to the power of

2

$2$ .

x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

Step 3.1.2

Multiply

- 4 \cdot 1 \cdot 10

$- 4 \cdot 1 \cdot 10$ .

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

1

$1$ .

x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 10}}{2 \cdot 1}

$x = \frac{- 7 \pm \sqrt{49 - 4 \cdot 10}}{2 \cdot 1}$

Step 3.1.2.2

Multiply

- 4

$- 4$ by

10

$10$ .

x = \frac{- 7 \pm \sqrt{49 - 40}}{2 \cdot 1}

$x = \frac{- 7 \pm \sqrt{49 - 40}}{2 \cdot 1}$

x = \frac{- 7 \pm \sqrt{49 - 40}}{2 \cdot 1}

$x = \frac{- 7 \pm \sqrt{49 - 40}}{2 \cdot 1}$

Step 3.1.3

Subtract

40

$40$ from

49

$49$ .

x = \frac{- 7 \pm \sqrt{9}}{2 \cdot 1}

$x = \frac{- 7 \pm \sqrt{9}}{2 \cdot 1}$

Step 3.1.4

Rewrite

$9$ as

$3^{2}$ .

$x = \frac{- 7 \pm \sqrt{3^{2}}}{2 \cdot 1}$

Step 3.1.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 7 \pm 3}{2 \cdot 1}$

Step 3.2

Multiply

$2$ by

$1$ .

$x = \frac{- 7 \pm 3}{2}$

Step 4

The final answer is the combination of both solutions.

$x = - 2, - 5$

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