Solve y^2+4y+8x-4=0

$$y ^ { 2 } + 4 y + 8 x - 4 = 0$$

Solve for x

$$x=-\frac{y^{2}}{8}-\frac{y}{2}+\frac{1}{2}$$

Steps for Solving Linear Equation

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

$$4y+8x-4=-y^{2}$$

Subtract 4y from both sides.

$$8x-4=-y^{2}-4y$$

Add 4 to both sides.

$$8x=-y^{2}-4y+4$$

The equation is in standard form.

$$8x=4-4y-y^{2}$$

Divide both sides by 8.

$$\frac{8x}{8}=\frac{4-4y-y^{2}}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x=\frac{4-4y-y^{2}}{8}$$

Divide -y^{2}-4y+4 by 8.

$$x=-\frac{y^{2}}{8}-\frac{y}{2}+\frac{1}{2}$$

Solve for y (complex solution)

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

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

Solve for y

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

$$y=-2\sqrt{2-2x}-2\text{, }x\leq 1$$