Solve Using the Quadratic Formula 2x^2-4x-16=0

$2 x^{2} - 4 x - 16 = 0$

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 $a = 2$ , $b = - 4$ , and $c = - 16$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

Raise $- 4$ to the power of $2$ .

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

Multiply $- 4 \cdot 2 \cdot - 16$ .

Multiply $- 4$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 8 \cdot - 16}}{2 \cdot 2}$

Multiply $- 8$ by $- 16$ .

$x = \frac{4 \pm \sqrt{16 + 128}}{2 \cdot 2}$

Add $16$ and $128$ .

$x = \frac{4 \pm \sqrt{144}}{2 \cdot 2}$

Rewrite $144$ as $12^{2}$ .

$x = \frac{4 \pm \sqrt{12^{2}}}{2 \cdot 2}$

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

$x = \frac{4 \pm 12}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{4 \pm 12}{4}$

Simplify $\frac{4 \pm 12}{4}$ .

$x = 1 \pm 3$

Step 4

The final answer is the combination of both solutions.

$x = 4, - 2$