Solve +4y^2+4x-8y+4=0

$$+ 4 y ^ { 2 } + 4 x - 8 y + 4 = 0$$

Solve for x

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

Steps for Solving Linear Equation

Subtract 4y^{2} from both sides. Anything subtracted from zero gives its negation.

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

Add 8y to both sides.

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

Subtract 4 from both sides.

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

Divide both sides by 4.

$$\frac{4x}{4}=-\frac{4\left(y-1\right)^{2}}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x=-\frac{4\left(y-1\right)^{2}}{4}$$

Divide -4\left(-1+y\right)^{2} by 4.

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

Solve for y (complex solution)

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

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

Solve for y

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

$$y=-\sqrt{-x}+1\text{, }x\leq 0$$