Solve x=-2+sqrt{4+4y-12}

$$x=-2+ \sqrt{ 4+4y-12 }$$

Solve for y

$$y=\frac{\left(x+2\right)^{2}+8}{4}$$

$$\frac{x}{2}+1\geq 0$$

Steps for Solving Radical Equations

Subtract 12 from 4 to get -8.

$$x=-2+\sqrt{-8+4y}$$

Swap sides so that all variable terms are on the left hand side.

$$-2+\sqrt{-8+4y}=x$$

Add 2 to both sides.

$$\sqrt{-8+4y}=x+2$$

Square both sides of the equation.

$$4y-8=\left(x+2\right)^{2}$$

Add 8 to both sides of the equation.

$$4y-8-\left(-8\right)=\left(x+2\right)^{2}-\left(-8\right)$$

Subtracting -8 from itself leaves 0.

$$4y=\left(x+2\right)^{2}-\left(-8\right)$$

Subtract -8 from \left(x+2\right)^{2}.

$$4y=\left(x+2\right)^{2}+8$$

Divide both sides by 4.

$$\frac{4y}{4}=\frac{\left(x+2\right)^{2}+8}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y=\frac{\left(x+2\right)^{2}+8}{4}$$

Divide \left(x+2\right)^{2}+8 by 4.

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

Solve for x (complex solution)

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

Solve for y (complex solution)

$$y=\frac{\left(x+2\right)^{2}+8}{4}$$

$$x=-2\text{ or }arg(\frac{x}{2}+1)<\pi$$

Solve for x

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

$$y\geq 2$$