Solve 16+left(8-xright)^2=x^2

$$16+ { \left(8-x \right) }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x=5$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$16+64-16x+x^{2}=x^{2}$$

Add 16 and 64 to get 80.

$$80-16x+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$80-16x+x^{2}-x^{2}=0$$

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

$$80-16x=0$$

Subtract 80 from both sides. Anything subtracted from zero gives its negation.

$$-16x=-80$$

Divide both sides by -16.

$$x=\frac{-80}{-16}$$

Divide -80 by -16 to get 5.

$$x=5$$