Solve 2x+3y^2=8x

$$2 x + 3 y ^ { 2 } = 8 x$$

Solve for x

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

Steps for Solving Linear Equation

Subtract 8x from both sides.

$$2x+3y^{2}-8x=0$$

Combine 2x and -8x to get -6x.

$$-6x+3y^{2}=0$$

Subtract 3y^{2} from both sides. Anything subtracted from zero gives its negation.

$$-6x=-3y^{2}$$

Divide both sides by -6.

$$\frac{-6x}{-6}=-\frac{3y^{2}}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x=-\frac{3y^{2}}{-6}$$

Divide -3y^{2} by -6.

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

Solve for y (complex solution)

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

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

Solve for y

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

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