Solve Using the Quadratic Formula 3x^2-4x=-2

$3 x^{2} - 4 x = - 2$

Step 1

Add $2$ to both sides of the equation.

$3 x^{2} - 4 x + 2 = 0$

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 $a = 3$ , $b = - 4$ , and $c = 2$ into the quadratic formula and solve for $x$ .

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

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 3 \cdot 2}}{2 \cdot 3}$

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

Multiply $- 4$ by $3$ .

$x = \frac{4 \pm \sqrt{16 - 12 \cdot 2}}{2 \cdot 3}$

Multiply $- 12$ by $2$ .

$x = \frac{4 \pm \sqrt{16 - 24}}{2 \cdot 3}$

Subtract $24$ from $16$ .

$x = \frac{4 \pm \sqrt{- 8}}{2 \cdot 3}$

Rewrite $- 8$ as $- 1 (8)$ .

$x = \frac{4 \pm \sqrt{- 1 \cdot 8}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$x = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot 3}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{4 \pm i \cdot \sqrt{8}}{2 \cdot 3}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

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

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

$x = \frac{4 \pm i \cdot \sqrt{2^{2} \cdot 2}}{2 \cdot 3}$

Pull terms out from under the radical.

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

Move $2$ to the left of $i$ .

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

Multiply $2$ by $3$ .

$x = \frac{4 \pm 2 i \sqrt{2}}{6}$

Simplify $\frac{4 \pm 2 i \sqrt{2}}{6}$ .

$x = \frac{2 \pm i \sqrt{2}}{3}$