Solve x=sqrt{y}+x^2

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

Solve for y

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

$$x-x^{2}\geq 0$$

Steps for Solving Radical Equations

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

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

Subtract x^{2} from both sides.

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

Square both sides of the equation.

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

Solve for x (complex solution)

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

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

Solve for y (complex solution)

$$\left\{\begin{matrix}y=\left(x\left(1-x\right)\right)^{2}\text{, }&arg(x-x^{2})<\pi \\y=0\text{, }&x=1\text{ or }x=0\end{matrix}\right.$$

Solve for x

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

$$x=\frac{\sqrt{-4\sqrt{y}+1}+1}{2}\text{, }y\geq 0\text{ and }y\leq \frac{1}{16}$$