Solve y=4-sqrt{5+x}

$$y=4- \sqrt{ 5+x }$$

Solve for x

$$x=\left(4-y\right)^{2}-5$$

$$4-y\geq 0$$

Steps for Solving Radical Equations

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

$$4-\sqrt{5+x}=y$$

Subtract 4 from both sides.

$$-\sqrt{5+x}=y-4$$

Divide both sides by -1.

$$\frac{-\sqrt{x+5}}{-1}=\frac{y-4}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$\sqrt{x+5}=\frac{y-4}{-1}$$

Divide y-4 by -1.

$$\sqrt{x+5}=4-y$$

Square both sides of the equation.

$$x+5=\left(4-y\right)^{2}$$

Subtract 5 from both sides of the equation.

$$x+5-5=\left(4-y\right)^{2}-5$$

Subtracting 5 from itself leaves 0.

$$x=\left(4-y\right)^{2}-5$$

Solve for x (complex solution)

$$x=\left(4-y\right)^{2}-5$$

$$y=4\text{ or }arg(4-y)<\pi$$

Solve for y (complex solution)

$$y=-\sqrt{x+5}+4$$

Solve for y

$$y=-\sqrt{x+5}+4$$

$$x\geq -5$$