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

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

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 $a = - 1$ , $b = - 4$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (- 1 \cdot 0)}}{2 \cdot - 1}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot - 1 \cdot 0}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 0$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{4 \pm \sqrt{16 + 4 \cdot 0}}{2 \cdot - 1}$

Multiply $4$ by $0$ .

$x = \frac{4 \pm \sqrt{16 + 0}}{2 \cdot - 1}$

Add $16$ and $0$ .

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

Rewrite $16$ as $4^{2}$ .

$x = \frac{4 \pm \sqrt{4^{2}}}{2 \cdot - 1}$

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

$x = \frac{4 \pm 4}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{4 \pm 4}{- 2}$

Simplify $\frac{4 \pm 4}{- 2}$ .

$x = \frac{2 \pm 2}{- 1}$

Move the negative one from the denominator of $\frac{2 \pm 2}{- 1}$ .

$x = - 1 \cdot (2 \pm 2)$

Rewrite $- 1 \cdot (2 \pm 2)$ as $- (2 \pm 2)$ .

$x = - (2 \pm 2)$

Step 4

The final answer is the combination of both solutions.

$x = - 4, 0$