Solve sqrt{x-3}+y^2+4y+4=0

$$\sqrt { x - 3 } + y ^ { 2 } + 4 y + 4 = 0$$

Solve for x

$$x=3$$

$$y=-2$$

Steps for Solving Radical Equations

Subtract y^{2}+4y+4 from both sides of the equation.

$$\sqrt{x-3}+y^{2}+4y+4-\left(y^{2}+4y+4\right)=-\left(y^{2}+4y+4\right)$$

Subtracting y^{2}+4y+4 from itself leaves 0.

$$\sqrt{x-3}=-\left(y^{2}+4y+4\right)$$

Subtract y^{2}+4y+4 from 0.

$$\sqrt{x-3}=-\left(y+2\right)^{2}$$

Square both sides of the equation.

$$x-3=\left(y+2\right)^{4}$$

Add 3 to both sides of the equation.

$$x-3-\left(-3\right)=\left(y+2\right)^{4}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$x=\left(y+2\right)^{4}-\left(-3\right)$$

Subtract -3 from \left(y+2\right)^{4}.

$$x=\left(y+2\right)^{4}+3$$

Solve for x (complex solution)

$$x=\left(y+2\right)^{4}+3$$

$$arg(\left(y+2\right)^{2})\geq \pi \text{ or }y=-2$$

Solve for y (complex solution)

$$y=\sqrt{-\sqrt{x-3}}-2$$

$$y=-\sqrt{-\sqrt{x-3}}-2$$

Solve for y

$$y=-2$$

$$x=3$$