Find All Complex Solutions -2y^2+4y+1=-6y^2

$- 2 y^{2} + 4 y + 1 = - 6 y^{2}$

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

Add $6 y^{2}$ to both sides of the equation.

$- 2 y^{2} + 4 y + 1 + 6 y^{2} = 0$

Add $- 2 y^{2}$ and $6 y^{2}$ .

$4 y^{2} + 4 y + 1 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 4$ , $b = 4$ , and $c = 1$ into the quadratic formula and solve for $y$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (4 \cdot 1)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

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

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$y = \frac{- 4 \pm \sqrt{16 - 16 \cdot 1}}{2 \cdot 4}$

Multiply $- 16$ by $1$ .

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

Subtract $16$ from $16$ .

$y = \frac{- 4 \pm \sqrt{0}}{2 \cdot 4}$

Rewrite $0$ as $0^{2}$ .

$y = \frac{- 4 \pm \sqrt{0^{2}}}{2 \cdot 4}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \frac{- 4 \pm 0}{2 \cdot 4}$

$- 4$ plus or minus $0$ is $- 4$ .

$y = \frac{- 4}{2 \cdot 4}$

Multiply $2$ by $4$ .

$y = \frac{- 4}{8}$

Cancel the common factor of $- 4$ and $8$ .

Factor $4$ out of $- 4$ .

$y = \frac{4 (- 1)}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

$y = \frac{- 1}{2}$

Move the negative in front of the fraction.

$y = - \frac{1}{2}$

The final answer is the combination of both solutions.

$y = - \frac{1}{2}$ Double roots