Solve y=sqrt{4-x}-4

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

Solve for x

$$x=-\left(y+2\right)\left(y+6\right)$$

$$y+4\geq 0$$

Steps for Solving Radical Equations

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

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

Add 4 to both sides.

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

Square both sides of the equation.

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

Subtract 4 from both sides of the equation.

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

Subtracting 4 from itself leaves 0.

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

Subtract 4 from \left(y+4\right)^{2}.

$$-x=\left(y+2\right)\left(y+6\right)$$

Divide both sides by -1.

$$\frac{-x}{-1}=\frac{\left(y+2\right)\left(y+6\right)}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x=\frac{\left(y+2\right)\left(y+6\right)}{-1}$$

Divide \left(2+y\right)\left(6+y\right) by -1.

$$x=-\left(y+2\right)\left(y+6\right)$$

Solve for x (complex solution)

$$x=-\left(y+2\right)\left(y+6\right)$$

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

Solve for y

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

$$x\leq 4$$