Solve 2x^2+3y-5=0

$$2 x ^ { 2 } + 3 y - 5 = 0$$

Solve for y

$$y=\frac{5-2x^{2}}{3}$$

Steps for Solving Linear Equation

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

$$3y-5=-2x^{2}$$

Add 5 to both sides.

$$3y=-2x^{2}+5$$

The equation is in standard form.

$$3y=5-2x^{2}$$

Divide both sides by 3.

$$\frac{3y}{3}=\frac{5-2x^{2}}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y=\frac{5-2x^{2}}{3}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{10-6y}}{2}$$

$$x=\frac{\sqrt{10-6y}}{2}$$

Solve for x

$$x=\frac{\sqrt{10-6y}}{2}$$

$$x=-\frac{\sqrt{10-6y}}{2}\text{, }y\leq \frac{5}{3}$$