Solve Using the Quadratic Formula 2x^2+6x=0

$2 x^{2} + 6 x = 0$

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 $a = 2$ , $b = 6$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (2 \cdot 0)}}{2 \cdot 2}$

Step 3

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 2 \cdot 0}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot 0$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 8 \cdot 0}}{2 \cdot 2}$

Multiply $- 8$ by $0$ .

$x = \frac{- 6 \pm \sqrt{36 + 0}}{2 \cdot 2}$

Add $36$ and $0$ .

$x = \frac{- 6 \pm \sqrt{36}}{2 \cdot 2}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 6 \pm \sqrt{6^{2}}}{2 \cdot 2}$

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

$x = \frac{- 6 \pm 6}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 6 \pm 6}{4}$

Simplify $\frac{- 6 \pm 6}{4}$ .

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

Step 4

The final answer is the combination of both solutions.

$x = 0, - 3$