Solve y=sqrt{3-x}+2

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

Solve for x

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

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

Steps for Solving Radical Equations

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

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

Subtract 2 from both sides.

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

Square both sides of the equation.

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

Subtract 3 from both sides of the equation.

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

Subtracting 3 from itself leaves 0.

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

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{\left(y-2\right)^{2}-3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{\left(y-2\right)^{2}-3}{-1}$$

Divide \left(y-2\right)^{2}-3 by -1.

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

Solve for x (complex solution)

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

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

Solve for y

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

$$x\leq 3$$