Solve y^2-4y+20x-16=0

$$y ^ { 2 } - 4 y + 20 x - 16 = 0$$

Solve for x

$$x=-\frac{y^{2}}{20}+\frac{y}{5}+\frac{4}{5}$$

Steps for Solving Linear Equation

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

$$-4y+20x-16=-y^{2}$$

Add 4y to both sides.

$$20x-16=-y^{2}+4y$$

Add 16 to both sides.

$$20x=-y^{2}+4y+16$$

The equation is in standard form.

$$20x=16+4y-y^{2}$$

Divide both sides by 20.

$$\frac{20x}{20}=\frac{16+4y-y^{2}}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x=\frac{16+4y-y^{2}}{20}$$

Divide -y^{2}+4y+16 by 20.

$$x=-\frac{y^{2}}{20}+\frac{y}{5}+\frac{4}{5}$$

Solve for y (complex solution)

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

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

Solve for y

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

$$y=-2\sqrt{5-5x}+2\text{, }x\leq 1$$