Solve Using the Quadratic Formula 6x^2+4x=0

$6 x^{2} + 4 x = 0$

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 $a = 6$ , $b = 4$ , and $c = 0$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 6 \cdot 0}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot 0$ .

Multiply $- 4$ by $6$ .

$x = \frac{- 4 \pm \sqrt{16 - 24 \cdot 0}}{2 \cdot 6}$

Multiply $- 24$ by $0$ .

$x = \frac{- 4 \pm \sqrt{16 + 0}}{2 \cdot 6}$

Add $16$ and $0$ .

$x = \frac{- 4 \pm \sqrt{16}}{2 \cdot 6}$

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

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

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

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

Multiply $2$ by $6$ .

$x = \frac{- 4 \pm 4}{12}$

Simplify $\frac{- 4 \pm 4}{12}$ .

$x = \frac{- 1 \pm 1}{3}$

Step 4

The final answer is the combination of both solutions.

$x = 0, - \frac{2}{3}$