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

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

Solve for y

$$y=\left(2x-3\right)\left(x+1\right)$$

Steps for Solving Linear Equation

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

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

Add x to both sides.

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

Add 3 to both sides.

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

The equation is in standard form.

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

Divide both sides by -1.

$$\frac{-y}{-1}=\frac{3+x-2x^{2}}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y=\frac{3+x-2x^{2}}{-1}$$

Divide -2x^{2}+x+3 by -1.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{8y+25}+1}{4}$$

$$x=\frac{-\sqrt{8y+25}+1}{4}$$

Solve for x

$$x=\frac{\sqrt{8y+25}+1}{4}$$

$$x=\frac{-\sqrt{8y+25}+1}{4}\text{, }y\geq -\frac{25}{8}$$