Solve (40-x)^2=x^2+400

$$( 40 - x ) ^ { 2 } = x ^ { 2 } + 400$$

Solve for x

$$x=15$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(40-x\right)^{2}.

$$1600-80x+x^{2}=x^{2}+400$$

Subtract x^{2} from both sides.

$$1600-80x+x^{2}-x^{2}=400$$

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

$$1600-80x=400$$

Subtract 1600 from both sides.

$$-80x=400-1600$$

Subtract 1600 from 400 to get -1200.

$$-80x=-1200$$

Divide both sides by -80.

$$x=\frac{-1200}{-80}$$

Divide -1200 by -80 to get 15.

$$x=15$$