Solve Using the Quadratic Formula x^2+9x+8=0

$x^{2} + 9 x + 8 = 0$

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 $a = 1$ , $b = 9$ , and $c = 8$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (1 \cdot 8)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot 8}}{2 \cdot 1}$

Multiply $- 4$ by $8$ .

$x = \frac{- 9 \pm \sqrt{81 - 32}}{2 \cdot 1}$

Subtract $32$ from $81$ .

$x = \frac{- 9 \pm \sqrt{49}}{2 \cdot 1}$

Rewrite $49$ as $7^{2}$ .

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

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

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

Multiply $2$ by $1$ .

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

Step 4

The final answer is the combination of both solutions.

$x = - 1, - 8$