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

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

Solve for y

$$y=-\frac{x^{2}}{4}-\frac{2x}{3}-\frac{2}{3}$$

$$x\geq 0$$

Steps for Solving Radical Equations

Add 2 and 0 to get 2.

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

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

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

Divide both sides by 2.

$$\frac{2\sqrt{3y+x^{2}+2x+2}}{2}=\frac{x}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$\sqrt{3y+x^{2}+2x+2}=\frac{x}{2}$$

Square both sides of the equation.

$$3y+x^{2}+2x+2=\frac{x^{2}}{4}$$

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

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

Subtracting 2+2x+x^{2} from itself leaves 0.

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

Subtract 2+2x+x^{2} from \frac{x^{2}}{4}.

$$3y=-\frac{3x^{2}}{4}-2x-2$$

Divide both sides by 3.

$$\frac{3y}{3}=\frac{-\frac{3x^{2}}{4}-2x-2}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y=\frac{-\frac{3x^{2}}{4}-2x-2}{3}$$

Divide -\frac{3x^{2}}{4}-2x-2 by 3.

$$y=-\frac{x^{2}}{4}-\frac{2x}{3}-\frac{2}{3}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{2\sqrt{-9y-2}-4}{3}\text{, }&|\frac{arg(-y-\frac{4\sqrt{-9y-2}}{9}+\frac{2}{9})}{2}-arg(\frac{2\sqrt{-9y-2}-4}{3})|<\pi \text{ or }y=-\frac{2}{3}\\x=\frac{-2\sqrt{-9y-2}-4}{3}\text{, }&|\frac{arg(-y+\frac{4\sqrt{-9y-2}}{9}+\frac{2}{9})}{2}-arg(\frac{-2\sqrt{-9y-2}-4}{3})|<\pi \end{matrix}\right.$$

Solve for y (complex solution)

$$y=-\frac{x^{2}}{4}-\frac{2x}{3}-\frac{2}{3}$$

$$arg(x)<\pi \text{ or }x=0$$

Solve for x

$$x=\frac{2\left(\sqrt{-9y-2}-2\right)}{3}$$

$$\left(\frac{2\sqrt{-9y-2}}{3}+\sqrt{-3y-1}\leq \frac{1}{3}\text{ or }\frac{2\sqrt{-9y-2}}{3}-\sqrt{-3y-1}\geq \frac{1}{3}\right)\text{ and }y\leq -\frac{2}{3}$$