Solve x^2-4x-8y+12=0

$$x ^ { 2 } - 4 x - 8 y + 12 = 0$$

Solve for y

$$y=\frac{x^{2}-4x+12}{8}$$

Steps for Solving Linear Equation

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

$$-4x-8y+12=-x^{2}$$

Add 4x to both sides.

$$-8y+12=-x^{2}+4x$$

Subtract 12 from both sides.

$$-8y=-x^{2}+4x-12$$

Divide both sides by -8.

$$\frac{-8y}{-8}=\frac{-x^{2}+4x-12}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$y=\frac{-x^{2}+4x-12}{-8}$$

Divide -x^{2}+4x-12 by -8.

$$y=\frac{x^{2}}{8}-\frac{x}{2}+\frac{3}{2}$$

Solve for x (complex solution)

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

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

Solve for x

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

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