Solve sqrt{5-a}=b+4

$$\sqrt { 5 - a } = b + 4$$

Solve for a

$$a=-\left(b+4\right)^{2}+5$$

$$b+4\geq 0$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$-a+5=\left(b+4\right)^{2}$$

Subtract 5 from both sides of the equation.

$$-a+5-5=\left(b+4\right)^{2}-5$$

Subtracting 5 from itself leaves 0.

$$-a=\left(b+4\right)^{2}-5$$

Divide both sides by -1.

$$\frac{-a}{-1}=\frac{\left(b+4\right)^{2}-5}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a=\frac{\left(b+4\right)^{2}-5}{-1}$$

Divide \left(b+4\right)^{2}-5 by -1.

$$a=-\left(b+4\right)^{2}+5$$

Solve for b

$$b=\sqrt{5-a}-4$$

$$a\leq 5$$