Solve 2x^2-5y-1=0

$$2 x ^ { 2 } - 5 y - 1 = 0$$

Solve for y

$$y=\frac{2x^{2}-1}{5}$$

Steps for Solving Linear Equation

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

$$-5y-1=-2x^{2}$$

Add 1 to both sides.

$$-5y=-2x^{2}+1$$

The equation is in standard form.

$$-5y=1-2x^{2}$$

Divide both sides by -5.

$$\frac{-5y}{-5}=\frac{1-2x^{2}}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$y=\frac{1-2x^{2}}{-5}$$

Divide -2x^{2}+1 by -5.

$$y=\frac{2x^{2}-1}{5}$$

Solve for x (complex solution)

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

$$x=\frac{\sqrt{10y+2}}{2}$$

Solve for x

$$x=\frac{\sqrt{10y+2}}{2}$$

$$x=-\frac{\sqrt{10y+2}}{2}\text{, }y\geq -\frac{1}{5}$$