Solve Using the Quadratic Formula 3x^2=2x+12

$3 x^{2} = 2 x + 12$

Step 1

Move all the expressions to the left side of the equation.

Subtract $2 x$ from both sides of the equation.

$3 x^{2} - 2 x = 12$

Subtract $12$ from both sides of the equation.

$3 x^{2} - 2 x - 12 = 0$

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 $a = 3$ , $b = - 2$ , and $c = - 12$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (3 \cdot - 12)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 3 \cdot - 12$ .

Multiply $- 4$ by $3$ .

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

Multiply $- 12$ by $- 12$ .

$x = \frac{2 \pm \sqrt{4 + 144}}{2 \cdot 3}$

Add $4$ and $144$ .

$x = \frac{2 \pm \sqrt{148}}{2 \cdot 3}$

Rewrite $148$ as $2^{2} \cdot 37$ .

Factor $4$ out of $148$ .

$x = \frac{2 \pm \sqrt{4 (37)}}{2 \cdot 3}$

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

$x = \frac{2 \pm \sqrt{2^{2} \cdot 37}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{37}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{2 \pm 2 \sqrt{37}}{6}$

Simplify $\frac{2 \pm 2 \sqrt{37}}{6}$ .

$x = \frac{1 \pm \sqrt{37}}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 \pm \sqrt{37}}{3}$

Decimal Form:

$x = 2.36092084 \dots, - 1.69425417 \dots$