Solve y^2-4x-5y+4=0

$$y ^ { 2 } - 4 x - 5 y + 4 = 0$$

Solve for x

$$x=\frac{\left(y-4\right)\left(y-1\right)}{4}$$

Steps for Solving Linear Equation

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

$$-4x-5y+4=-y^{2}$$

Add 5y to both sides.

$$-4x+4=-y^{2}+5y$$

Subtract 4 from both sides.

$$-4x=-y^{2}+5y-4$$

Divide both sides by -4.

$$\frac{-4x}{-4}=\frac{\left(1-y\right)\left(y-4\right)}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x=\frac{\left(1-y\right)\left(y-4\right)}{-4}$$

Divide \left(-4+y\right)\left(1-y\right) by -4.

$$x=-\frac{\left(1-y\right)\left(y-4\right)}{4}$$

Solve for y (complex solution)

$$y=\frac{\sqrt{16x+9}+5}{2}$$

$$y=\frac{-\sqrt{16x+9}+5}{2}$$

Solve for y

$$y=\frac{\sqrt{16x+9}+5}{2}$$

$$y=\frac{-\sqrt{16x+9}+5}{2}\text{, }x\geq -\frac{9}{16}$$