Solve for x 3x^2+5=-2x

$3 x^{2} + 5 = - 2 x$

Step 1

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

$3 x^{2} + 5 + 2 x = 0$

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 $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (3 \cdot 5)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 5$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 2 \pm \sqrt{4 - 12 \cdot 5}}{2 \cdot 3}$

Multiply $- 12$ by $5$ .

$x = \frac{- 2 \pm \sqrt{4 - 60}}{2 \cdot 3}$

Subtract $60$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 56}}{2 \cdot 3}$

Rewrite $- 56$ as $- 1 (56)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 56}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (56)}$ as $\sqrt{- 1} \cdot \sqrt{56}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{56}}{2 \cdot 3}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{56}}{2 \cdot 3}$

Rewrite $56$ as $2^{2} \cdot 14$ .

Factor $4$ out of $56$ .

$x = \frac{- 2 \pm i \cdot \sqrt{4 (14)}}{2 \cdot 3}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{2^{2} \cdot 14}}{2 \cdot 3}$

Pull terms out from under the radical.

$x = \frac{- 2 \pm i \cdot (2 \sqrt{14})}{2 \cdot 3}$

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

$x = \frac{- 2 \pm 2 i \sqrt{14}}{2 \cdot 3}$

Multiply $2$ by $3$ .

$x = \frac{- 2 \pm 2 i \sqrt{14}}{6}$

Simplify $\frac{- 2 \pm 2 i \sqrt{14}}{6}$ .

$x = \frac{- 1 \pm i \sqrt{14}}{3}$

Step 5

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{14}}{3}, - \frac{1 + i \sqrt{14}}{3}$