Solve 2x^2-3y+10=0

$$2 { x }^{ 2 } -3y+10=0$$

Solve for y

$$y=\frac{2\left(x^{2}+5\right)}{3}$$

Steps for Solving Linear Equation

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

$$-3y+10=-2x^{2}$$

Subtract 10 from both sides.

$$-3y=-2x^{2}-10$$

Divide both sides by -3.

$$\frac{-3y}{-3}=\frac{-2x^{2}-10}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$y=\frac{-2x^{2}-10}{-3}$$

Divide -2x^{2}-10 by -3.

$$y=\frac{2x^{2}+10}{3}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{6y-20}}{2}$$

$$x=\frac{\sqrt{6y-20}}{2}$$

Solve for x

$$x=\frac{\sqrt{6y-20}}{2}$$

$$x=-\frac{\sqrt{6y-20}}{2}\text{, }y\geq \frac{10}{3}$$