Solve for x 3x^2=6-2x

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

Step 1

Add $2 x$ to both sides of the equation.

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

Step 2

Subtract $6$ from both sides of the equation.

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

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

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

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

Step 5

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 3 \cdot - 6$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 2 \pm \sqrt{4 - 12 \cdot - 6}}{2 \cdot 3}$

Multiply $- 12$ by $- 6$ .

$x = \frac{- 2 \pm \sqrt{4 + 72}}{2 \cdot 3}$

Add $4$ and $72$ .

$x = \frac{- 2 \pm \sqrt{76}}{2 \cdot 3}$

Rewrite $76$ as $2^{2} \cdot 19$ .

Factor $4$ out of $76$ .

$x = \frac{- 2 \pm \sqrt{4 (19)}}{2 \cdot 3}$

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

$x = \frac{- 2 \pm \sqrt{2^{2} \cdot 19}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{19}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 2 \pm 2 \sqrt{19}}{6}$

Simplify $\frac{- 2 \pm 2 \sqrt{19}}{6}$ .

$x = \frac{- 1 \pm \sqrt{19}}{3}$

Step 6

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{19}}{3}, - \frac{1 + \sqrt{19}}{3}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{1 - \sqrt{19}}{3}, - \frac{1 + \sqrt{19}}{3}$

Decimal Form:

$x = 1.11963298 \dots, - 1.78629964 \dots$