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

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

Solve for y

$$y=-\frac{x^{2}}{4}+2x+1$$

Steps for Solving Linear Equation

Add 12y to both sides.

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

Combine -8y and 12y to get 4y.

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

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

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

Add 8x to both sides.

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

Add 4 to both sides.

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

The equation is in standard form.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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

$$y=-\frac{x^{2}}{4}+2x+1$$

Solve for x (complex solution)

$$x=2\sqrt{5-y}+4$$

$$x=-2\sqrt{5-y}+4$$

Solve for x

$$x=2\sqrt{5-y}+4$$

$$x=-2\sqrt{5-y}+4\text{, }y\leq 5$$