Solve y=2sqrt{-x}+3

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

Solve for x

$$x=-\frac{\left(3-y\right)^{2}}{4}$$

$$-\frac{3-y}{2}\geq 0$$

Steps for Solving Radical Equations

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

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

Subtract 3 from both sides.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Square both sides of the equation.

$$-x=\frac{\left(y-3\right)^{2}}{4}$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{\left(y-3\right)^{2}}{-4}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{\left(y-3\right)^{2}}{-4}$$

Divide \frac{\left(y-3\right)^{2}}{4} by -1.

$$x=-\frac{\left(y-3\right)^{2}}{4}$$

Solve for x (complex solution)

$$x=-\frac{\left(3-y\right)^{2}}{4}$$

$$y=3\text{ or }arg(\frac{3-y}{2})\geq \pi$$

Solve for y

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

$$x\leq 0$$