Solve Using the Quadratic Formula 7x^2+14x=0

$7 x^{2} + 14 x = 0$

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 $a = 7$ , $b = 14$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 14 \pm \sqrt{14^{2} - 4 \cdot (7 \cdot 0)}}{2 \cdot 7}$

Step 3

Simplify.

Simplify the numerator.

Raise $14$ to the power of $2$ .

$x = \frac{- 14 \pm \sqrt{196 - 4 \cdot 7 \cdot 0}}{2 \cdot 7}$

Multiply $- 4 \cdot 7 \cdot 0$ .

Multiply $- 4$ by $7$ .

$x = \frac{- 14 \pm \sqrt{196 - 28 \cdot 0}}{2 \cdot 7}$

Multiply $- 28$ by $0$ .

$x = \frac{- 14 \pm \sqrt{196 + 0}}{2 \cdot 7}$

Add $196$ and $0$ .

$x = \frac{- 14 \pm \sqrt{196}}{2 \cdot 7}$

Rewrite $196$ as $14^{2}$ .

$x = \frac{- 14 \pm \sqrt{14^{2}}}{2 \cdot 7}$

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

$x = \frac{- 14 \pm 14}{2 \cdot 7}$

Multiply $2$ by $7$ .

$x = \frac{- 14 \pm 14}{14}$

Simplify $\frac{- 14 \pm 14}{14}$ .

$x = - 1 \pm 1$

Step 4

The final answer is the combination of both solutions.

$x = 0, - 2$