Solve x^2=(x-2)^2+20^2

$$x ^ { 2 } = ( x - 2 ) ^ { 2 } + 20 ^ { 2 }$$

Solve for x

$$x=101$$

Steps for Solving Linear Equation

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

$$x^{2}=x^{2}-4x+4+20^{2}$$

Calculate 20 to the power of 2 and get 400.

$$x^{2}=x^{2}-4x+4+400$$

Add 4 and 400 to get 404.

$$x^{2}=x^{2}-4x+404$$

Subtract x^{2} from both sides.

$$x^{2}-x^{2}=-4x+404$$

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

$$0=-4x+404$$

Swap sides so that all variable terms are on the left hand side.

$$-4x+404=0$$

Subtract 404 from both sides. Anything subtracted from zero gives its negation.

$$-4x=-404$$

Divide both sides by -4.

$$x=\frac{-404}{-4}$$

Divide -404 by -4 to get 101.

$$x=101$$