Solve Using the Quadratic Formula 3x^2+18x=0

$3 x^{2} + 18 x = 0$

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 $a = 3$ , $b = 18$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 18 \pm \sqrt{18^{2} - 4 \cdot (3 \cdot 0)}}{2 \cdot 3}$

Step 3

Simplify.

Simplify the numerator.

Raise $18$ to the power of $2$ .

$x = \frac{- 18 \pm \sqrt{324 - 4 \cdot 3 \cdot 0}}{2 \cdot 3}$

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

Multiply $- 4$ by $3$ .

$x = \frac{- 18 \pm \sqrt{324 - 12 \cdot 0}}{2 \cdot 3}$

Multiply $- 12$ by $0$ .

$x = \frac{- 18 \pm \sqrt{324 + 0}}{2 \cdot 3}$

Add $324$ and $0$ .

$x = \frac{- 18 \pm \sqrt{324}}{2 \cdot 3}$

Rewrite $324$ as $18^{2}$ .

$x = \frac{- 18 \pm \sqrt{18^{2}}}{2 \cdot 3}$

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

$x = \frac{- 18 \pm 18}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 18 \pm 18}{6}$

Simplify $\frac{- 18 \pm 18}{6}$ .

$x = - 3 \pm 3$

Step 4

The final answer is the combination of both solutions.

$x = 0, - 6$