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

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

Solve for x

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

Steps for Solving Linear Equation

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

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

Add 8y to both sides.

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

Subtract 8 from both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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

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

Solve for y (complex solution)

$$y=2\sqrt{2-x}+4$$

$$y=-2\sqrt{2-x}+4$$

Solve for y

$$y=2\sqrt{2-x}+4$$

$$y=-2\sqrt{2-x}+4\text{, }x\leq 2$$