Solve Using the Quadratic Formula x^2+16x+55=0

$x^{2} + 16 x + 55 = 0$

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 $a = 1$ , $b = 16$ , and $c = 55$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (1 \cdot 55)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 1 \cdot 55}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 55$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 55}}{2 \cdot 1}$

Multiply $- 4$ by $55$ .

$x = \frac{- 16 \pm \sqrt{256 - 220}}{2 \cdot 1}$

Subtract $220$ from $256$ .

$x = \frac{- 16 \pm \sqrt{36}}{2 \cdot 1}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 16 \pm \sqrt{6^{2}}}{2 \cdot 1}$

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

$x = \frac{- 16 \pm 6}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 16 \pm 6}{2}$

Simplify $\frac{- 16 \pm 6}{2}$ .

$x = - 8 \pm 3$

Step 4

The final answer is the combination of both solutions.

$x = - 5, - 11$