Solve r^2=left(r-10right)^2+20^2

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

Solve for r

$$r=25$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(r-10\right)^{2}.

$$r^{2}=r^{2}-20r+100+20^{2}$$

Calculate 20 to the power of 2 and get 400.

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

Add 100 and 400 to get 500.

$$r^{2}=r^{2}-20r+500$$

Subtract r^{2} from both sides.

$$r^{2}-r^{2}=-20r+500$$

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

$$0=-20r+500$$

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

$$-20r+500=0$$

Subtract 500 from both sides. Anything subtracted from zero gives its negation.

$$-20r=-500$$

Divide both sides by -20.

$$r=\frac{-500}{-20}$$

Divide -500 by -20 to get 25.

$$r=25$$