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

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

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 $a = 2$ , $b = 16$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (2 \cdot 0)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 16 \pm \sqrt{256 - 4 \cdot 2 \cdot 0}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 0$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 16 \pm \sqrt{256 - 8 \cdot 0}}{2 \cdot 2}$

Multiply $- 8$ by $0$ .

$x = \frac{- 16 \pm \sqrt{256 + 0}}{2 \cdot 2}$

Add $256$ and $0$ .

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

Rewrite $256$ as $16^{2}$ .

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

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

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

Multiply $2$ by $2$ .

$x = \frac{- 16 \pm 16}{4}$

Simplify $\frac{- 16 \pm 16}{4}$ .

$x = - 4 \pm 4$

Step 4

The final answer is the combination of both solutions.

$x = 0, - 8$