Solve left(10+xright)^2=20^2+x^2

$${ \left(10+x \right) }^{ 2 } = { 20 }^{ 2 } + { x }^{ 2 }$$

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(10+x\right)^{2}.

$$100+20x+x^{2}=20^{2}+x^{2}$$

Calculate 20 to the power of 2 and get 400.

$$100+20x+x^{2}=400+x^{2}$$

Subtract x^{2} from both sides.

$$100+20x+x^{2}-x^{2}=400$$

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

$$100+20x=400$$

Subtract 100 from both sides.

$$20x=400-100$$

Subtract 100 from 400 to get 300.

$$20x=300$$

Divide both sides by 20.

$$x=\frac{300}{20}$$

Divide 300 by 20 to get 15.

$$x=15$$