Solve 2x^2-4y+3z=0

$$2 x ^ { 2 } - 4 y + 3 z = 0$$

Solve for y

$$y=\frac{x^{2}}{2}+\frac{3z}{4}$$

Steps for Solving Linear Equation

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

$$-4y+3z=-2x^{2}$$

Subtract 3z from both sides.

$$-4y=-2x^{2}-3z$$

Divide both sides by -4.

$$\frac{-4y}{-4}=\frac{-2x^{2}-3z}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$y=\frac{-2x^{2}-3z}{-4}$$

Divide -2x^{2}-3z by -4.

$$y=\frac{x^{2}}{2}+\frac{3z}{4}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{8y-6z}}{2}$$

$$x=\frac{\sqrt{8y-6z}}{2}$$

Solve for x

$$x=\frac{\sqrt{8y-6z}}{2}$$

$$x=-\frac{\sqrt{8y-6z}}{2}\text{, }y\geq \frac{3z}{4}$$