Solve sqrt{x-2}+y^2=y-1/4

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

Solve for x

$$x=\frac{\left(2y-1\right)^{4}+32}{16}$$

$$-y^{2}+y-\frac{1}{4}\geq 0$$

Steps for Solving Radical Equations

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

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

Subtracting y^{2} from itself leaves 0.

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

Subtract y^{2} from y-\frac{1}{4}.

$$\sqrt{x-2}=-\frac{\left(2y-1\right)^{2}}{4}$$

Square both sides of the equation.

$$x-2=\frac{\left(2y-1\right)^{4}}{16}$$

Add 2 to both sides of the equation.

$$x-2-\left(-2\right)=\frac{\left(2y-1\right)^{4}}{16}-\left(-2\right)$$

Subtracting -2 from itself leaves 0.

$$x=\frac{\left(2y-1\right)^{4}}{16}-\left(-2\right)$$

Subtract -2 from \frac{\left(2y-1\right)^{4}}{16}.

$$x=\frac{\left(2y-1\right)^{4}}{16}+2$$

Solve for x (complex solution)

$$x=\frac{\left(2y-1\right)^{4}+32}{16}$$

$$y=\frac{1}{2}\text{ or }arg(y^{2}-y+\frac{1}{4})\geq \pi$$

Solve for y (complex solution)

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

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

Solve for y

$$y = \frac{1}{2} = 0.5$$

$$x=2$$