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

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

Solve for x

$$x=\frac{\left(y-2\right)^{2}}{8}$$

Steps for Solving Linear Equation

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

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

Add 4y to both sides.

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

Subtract 4 from both sides.

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

Divide both sides by -8.

$$\frac{-8x}{-8}=-\frac{\left(y-2\right)^{2}}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x=-\frac{\left(y-2\right)^{2}}{-8}$$

Divide -\left(y-2\right)^{2} by -8.

$$x=\frac{\left(y-2\right)^{2}}{8}$$

Solve for y (complex solution)

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

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

Solve for y

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

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