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

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

Solve for y

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

$$x\neq -3$$

Steps for Solving Linear Equation

Add x^{2} to both sides. Anything plus zero gives itself.

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

Add 2x to both sides.

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

Combine all terms containing y.

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

Divide both sides by x+3.

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

Dividing by x+3 undoes the multiplication by x+3.

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{y^{2}+8y+4}+y-2}{2}$$

$$x=\frac{\sqrt{y^{2}+8y+4}+y-2}{2}$$

Solve for x

$$x=\frac{-\sqrt{y^{2}+8y+4}+y-2}{2}$$

$$x=\frac{\sqrt{y^{2}+8y+4}+y-2}{2}\text{, }y\geq 2\sqrt{3}-4\text{ or }y\leq -2\sqrt{3}-4$$