Find All Complex Solutions 3y^2+7y+3=-y^2

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

Step 1

Move all terms to the left side of the equation and simplify.

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

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

Add $3 y^{2}$ and $y^{2}$ .

$4 y^{2} + 7 y + 3 = 0$

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 $a = 4$ , $b = 7$ , and $c = 3$ into the quadratic formula and solve for $y$ .

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

Step 4

Simplify.

Simplify the numerator.

Raise $7$ to the power of $2$ .

$y = \frac{- 7 \pm \sqrt{49 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

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

Multiply $- 4$ by $4$ .

$y = \frac{- 7 \pm \sqrt{49 - 16 \cdot 3}}{2 \cdot 4}$

Multiply $- 16$ by $3$ .

$y = \frac{- 7 \pm \sqrt{49 - 48}}{2 \cdot 4}$

Subtract $48$ from $49$ .

$y = \frac{- 7 \pm \sqrt{1}}{2 \cdot 4}$

Any root of $1$ is $1$ .

$y = \frac{- 7 \pm 1}{2 \cdot 4}$

Multiply $2$ by $4$ .

$y = \frac{- 7 \pm 1}{8}$

Step 5

The final answer is the combination of both solutions.

$y = - \frac{3}{4}, - 1$