Solve sqrt{x-3}+2=y

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

Solve for x

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

$$-\left(2-y\right)\geq 0$$

Steps for Solving Radical Equations

Subtract 2 from both sides of the equation.

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

Subtracting 2 from itself leaves 0.

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

Square both sides of the equation.

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

Add 3 to both sides of the equation.

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

Subtracting -3 from itself leaves 0.

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

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

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

Solve for x (complex solution)

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

$$y=2\text{ or }arg(2-y)\geq \pi$$

Solve for y (complex solution)

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

Solve for y

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

$$x\geq 3$$