Solve 8x^2+12y=0

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

Solve for y

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

Steps for Solving Linear Equation

Subtract 8x^{2} from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by 12.

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

Dividing by 12 undoes the multiplication by 12.

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

Divide -8x^{2} by 12.

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

Solve for x (complex solution)

$$x=-\frac{i\sqrt{6y}}{2}$$

$$x=\frac{i\sqrt{6y}}{2}$$

Solve for x

$$x=\frac{\sqrt{-6y}}{2}$$

$$x=-\frac{\sqrt{-6y}}{2}\text{, }y\leq 0$$