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

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

Solve for x

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

Steps for Solving Linear Equation

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

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

Add 2y to both sides.

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

Subtract 5 from both sides.

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

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

Divide -y^{2}+2y-5 by 4.

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

Solve for y (complex solution)

$$y=2\sqrt{-x-1}+1$$

$$y=-2\sqrt{-x-1}+1$$

Solve for y

$$y=2\sqrt{-x-1}+1$$

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