Solve Using the Quadratic Formula 3x^2+12x=0

$3 x^{2} + 12 x = 0$

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 $a = 3$ , $b = 12$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (3 \cdot 0)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 3 \cdot 0}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 0$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 12 \pm \sqrt{144 - 12 \cdot 0}}{2 \cdot 3}$

Multiply $- 12$ by $0$ .

$x = \frac{- 12 \pm \sqrt{144 + 0}}{2 \cdot 3}$

Add $144$ and $0$ .

$x = \frac{- 12 \pm \sqrt{144}}{2 \cdot 3}$

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

$x = \frac{- 12 \pm \sqrt{12^{2}}}{2 \cdot 3}$

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

$x = \frac{- 12 \pm 12}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 12 \pm 12}{6}$

Simplify $\frac{- 12 \pm 12}{6}$ .

$x = - 2 \pm 2$

Step 4

The final answer is the combination of both solutions.

$x = 0, - 4$