Solve Using the Quadratic Formula 5x^2+20x=0

$5 x^{2} + 20 x = 0$

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 $a = 5$ , $b = 20$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 20 \pm \sqrt{20^{2} - 4 \cdot (5 \cdot 0)}}{2 \cdot 5}$

Step 3

Simplify.

Simplify the numerator.

Raise $20$ to the power of $2$ .

$x = \frac{- 20 \pm \sqrt{400 - 4 \cdot 5 \cdot 0}}{2 \cdot 5}$

Multiply $- 4 \cdot 5 \cdot 0$ .

Multiply $- 4$ by $5$ .

$x = \frac{- 20 \pm \sqrt{400 - 20 \cdot 0}}{2 \cdot 5}$

Multiply $- 20$ by $0$ .

$x = \frac{- 20 \pm \sqrt{400 + 0}}{2 \cdot 5}$

Add $400$ and $0$ .

$x = \frac{- 20 \pm \sqrt{400}}{2 \cdot 5}$

Rewrite $400$ as $20^{2}$ .

$x = \frac{- 20 \pm \sqrt{20^{2}}}{2 \cdot 5}$

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

$x = \frac{- 20 \pm 20}{2 \cdot 5}$

Multiply $2$ by $5$ .

$x = \frac{- 20 \pm 20}{10}$

Simplify $\frac{- 20 \pm 20}{10}$ .

$x = - 2 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = 0, - 4$