Solve Using the Quadratic Formula x^2-7x-8=0

$x^{2} - 7 x - 8 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 7$ , and $c = - 8$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (1 \cdot - 8)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 1 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 8$ .

Multiply $- 4$ by $1$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot - 8}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$x = \frac{7 \pm \sqrt{49 + 32}}{2 \cdot 1}$

Add $49$ and $32$ .

$x = \frac{7 \pm \sqrt{81}}{2 \cdot 1}$

Rewrite $81$ as $9^{2}$ .

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

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

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

Multiply $2$ by $1$ .

$x = \frac{7 \pm 9}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 8, - 1$