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

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

Solve for x

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

Steps for Solving Linear Equation

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

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

Subtract 16y from both sides.

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

Subtract 4 from both sides.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

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

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

$$x=\frac{y^{2}}{4}+4y+1$$

Solve for y (complex solution)

$$y=2\sqrt{x+15}-8$$

$$y=-2\sqrt{x+15}-8$$

Solve for y

$$y=2\sqrt{x+15}-8$$

$$y=-2\sqrt{x+15}-8\text{, }x\geq -15$$