Solve 2x^2+8x-y+8=0

$$2 x ^ { 2 } + 8 x - y + 8 = 0$$

Solve for y

$$y=2\left(x+2\right)^{2}$$

Steps for Solving Linear Equation

Subtract 2x^{2} from both sides. Anything subtracted from zero gives its negation.

$$8x-y+8=-2x^{2}$$

Subtract 8x from both sides.

$$-y+8=-2x^{2}-8x$$

Subtract 8 from both sides.

$$-y=-2x^{2}-8x-8$$

Divide both sides by -1.

$$\frac{-y}{-1}=-\frac{2\left(x+2\right)^{2}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y=-\frac{2\left(x+2\right)^{2}}{-1}$$

Divide -2\left(2+x\right)^{2} by -1.

$$y=2\left(x+2\right)^{2}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{2y}}{2}-2$$

$$x=-\frac{\sqrt{2y}}{2}-2$$

Solve for x

$$x=\frac{\sqrt{2y}}{2}-2$$

$$x=-\frac{\sqrt{2y}}{2}-2\text{, }y\geq 0$$