Solve y=-sqrt{x-2}+3

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

Solve for x

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

$$3-y\geq 0$$

Steps for Solving Radical Equations

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

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

Subtract 3 from both sides.

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

Divide both sides by -1.

$$\frac{-\sqrt{x-2}}{-1}=\frac{y-3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$\sqrt{x-2}=\frac{y-3}{-1}$$

Divide y-3 by -1.

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

Square both sides of the equation.

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

Add 2 to both sides of the equation.

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

Subtracting -2 from itself leaves 0.

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

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

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

Solve for x (complex solution)

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

$$y=3\text{ or }arg(3-y)<\pi$$

Solve for y

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

$$x\geq 2$$