Solve Using the Quadratic Formula 3x^2+6x=2

$3 x^{2} + 6 x = 2$

Step 1

Subtract $2$ from both sides of the equation.

$3 x^{2} + 6 x - 2 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 3$ , $b = 6$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (3 \cdot - 2)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 3 \cdot - 2}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot - 2$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 6 \pm \sqrt{36 - 12 \cdot - 2}}{2 \cdot 3}$

Multiply $- 12$ by $- 2$ .

$x = \frac{- 6 \pm \sqrt{36 + 24}}{2 \cdot 3}$

Add $36$ and $24$ .

$x = \frac{- 6 \pm \sqrt{60}}{2 \cdot 3}$

Rewrite $60$ as $2^{2} \cdot 15$ .

Factor $4$ out of $60$ .

$x = \frac{- 6 \pm \sqrt{4 (15)}}{2 \cdot 3}$

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

$x = \frac{- 6 \pm \sqrt{2^{2} \cdot 15}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm 2 \sqrt{15}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 6 \pm 2 \sqrt{15}}{6}$

Simplify $\frac{- 6 \pm 2 \sqrt{15}}{6}$ .

$x = \frac{- 3 \pm \sqrt{15}}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 3 \pm \sqrt{15}}{3}$

Decimal Form:

$x = 0.29099444 \dots, - 2.29099444 \dots$