Solve sqrt{{left(-b/2right)}^2+x}=4

$$\sqrt{ { \left(- \frac{ b }{ 2 } \right) }^{ 2 } +x } = 4$$

Solve for x

$$x=-\frac{b^{2}}{4}+16$$

Steps for Solving Radical Equations

Square both sides of the equation.

$$x+\frac{b^{2}}{4}=16$$

Subtract \frac{1}{4}b^{2} from both sides of the equation.

$$x+\frac{b^{2}}{4}-\frac{b^{2}}{4}=16-\frac{b^{2}}{4}$$

Subtracting \frac{1}{4}b^{2} from itself leaves 0.

$$x=16-\frac{b^{2}}{4}$$

Subtract \frac{1}{4}b^{2} from 16.

$$x=-\frac{b^{2}}{4}+16$$

Solve for b (complex solution)

$$b=-2\sqrt{16-x}$$

$$b=2\sqrt{16-x}$$

Solve for b

$$b=2\sqrt{16-x}$$

$$b=-2\sqrt{16-x}\text{, }x\leq 16$$