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

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

Solve for y

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

Steps for Solving Linear Equation

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

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

Add 3x to both sides.

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

Subtract 2 from both sides.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{8y-7}+3}{4}$$

$$x=\frac{-\sqrt{8y-7}+3}{4}$$

Solve for x

$$x=\frac{\sqrt{8y-7}+3}{4}$$

$$x=\frac{-\sqrt{8y-7}+3}{4}\text{, }y\geq \frac{7}{8}$$