Find the Roots (Zeros) 3x^2+2x=-7

$3 x^{2} + 2 x = - 7$

Step 1

Add $7$ to both sides of the equation.

$3 x^{2} + 2 x + 7 = 0$

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 $c = 7$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (3 \cdot 7)}}{2 \cdot 3}$

Step 4

Simplify.

Simplify the numerator.

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 3 \cdot 7}}{2 \cdot 3}$

Multiply $- 4 \cdot 3 \cdot 7$ .

Multiply $- 4$ by $3$ .

$x = \frac{- 2 \pm \sqrt{4 - 12 \cdot 7}}{2 \cdot 3}$

Multiply $- 12$ by $7$ .

$x = \frac{- 2 \pm \sqrt{4 - 84}}{2 \cdot 3}$

Subtract $84$ from $4$ .

$x = \frac{- 2 \pm \sqrt{- 80}}{2 \cdot 3}$

Rewrite $- 80$ as $- 1 (80)$ .

$x = \frac{- 2 \pm \sqrt{- 1 \cdot 80}}{2 \cdot 3}$

Rewrite $\sqrt{- 1 (80)}$ as $\sqrt{- 1} \cdot \sqrt{80}$ .

$x = \frac{- 2 \pm \sqrt{- 1} \cdot \sqrt{80}}{2 \cdot 3}$

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

$x = \frac{- 2 \pm i \cdot \sqrt{80}}{2 \cdot 3}$

Rewrite $80$ as $4^{2} \cdot 5$ .

Factor $16$ out of $80$ .

$x = \frac{- 2 \pm i \cdot \sqrt{16 (5)}}{2 \cdot 3}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

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

Move $4$ to the left of $i$ .

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

Multiply $2$ by $3$ .

$x = \frac{- 2 \pm 4 i \sqrt{5}}{6}$

Simplify $\frac{- 2 \pm 4 i \sqrt{5}}{6}$ .

$x = \frac{- 1 \pm 2 i \sqrt{5}}{3}$

Step 5