Solve left(4-zright)^2=z^2+4

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

Solve for z

$$z = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-z\right)^{2}.

$$16-8z+z^{2}=z^{2}+4$$

Subtract z^{2} from both sides.

$$16-8z+z^{2}-z^{2}=4$$

Combine z^{2} and -z^{2} to get 0.

$$16-8z=4$$

Subtract 16 from both sides.

$$-8z=4-16$$

Subtract 16 from 4 to get -12.

$$-8z=-12$$

Divide both sides by -8.

$$z=\frac{-12}{-8}$$

Reduce the fraction \frac{-12}{-8} to lowest terms by extracting and canceling out -4.

$$z=\frac{3}{2}$$