Solve 2x^2-xy-3x+y+2012=0

$$2 x ^ { 2 } - x y - 3 x + y + 2012 = 0$$

Solve for y

$$y=-\frac{2x^{2}-3x+2012}{1-x}$$

$$x\neq 1$$

Steps for Solving Linear Equation

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

$$-xy-3x+y+2012=-2x^{2}$$

Add 3x to both sides.

$$-xy+y+2012=-2x^{2}+3x$$

Subtract 2012 from both sides.

$$-xy+y=-2x^{2}+3x-2012$$

Combine all terms containing y.

$$\left(-x+1\right)y=-2x^{2}+3x-2012$$

The equation is in standard form.

$$\left(1-x\right)y=-2x^{2}+3x-2012$$

Divide both sides by -x+1.

$$\frac{\left(1-x\right)y}{1-x}=\frac{-2x^{2}+3x-2012}{1-x}$$

Dividing by -x+1 undoes the multiplication by -x+1.

$$y=\frac{-2x^{2}+3x-2012}{1-x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}-2y-16087}+y+3}{4}$$

$$x=\frac{-\sqrt{y^{2}-2y-16087}+y+3}{4}$$

Solve for x

$$x=\frac{\sqrt{y^{2}-2y-16087}+y+3}{4}$$

$$x=\frac{-\sqrt{y^{2}-2y-16087}+y+3}{4}\text{, }y\geq 2\sqrt{4022}+1\text{ or }y\leq 1-2\sqrt{4022}$$