Solve Using the Quadratic Formula 6y+3=-4y^2

$6 y + 3 = - 4 y^{2}$

Step 1

Add $4 y^{2}$ to both sides of the equation.

$6 y + 3 + 4 y^{2} = 0$

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 $a = 4$ , $b = 6$ , and $c = 3$ into the quadratic formula and solve for $y$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$y = \frac{- 6 \pm \sqrt{36 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$y = \frac{- 6 \pm \sqrt{36 - 16 \cdot 3}}{2 \cdot 4}$

Multiply $- 16$ by $3$ .

$y = \frac{- 6 \pm \sqrt{36 - 48}}{2 \cdot 4}$

Subtract $48$ from $36$ .

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

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{- 6 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

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

$y = \frac{- 6 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$y = \frac{- 6 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

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

$y = \frac{- 6 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

$y = \frac{- 6 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

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

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

Multiply $2$ by $4$ .

$y = \frac{- 6 \pm 2 i \sqrt{3}}{8}$

Simplify $\frac{- 6 \pm 2 i \sqrt{3}}{8}$ .

$y = \frac{- 3 \pm i \sqrt{3}}{4}$