Solve r^2=6+(r-2)^2

$$r ^ { 2 } = 6 + ( r - 2 ) ^ { 2 }$$

Solve for r

$$r = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps for Solving Linear Equation

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

$$r^{2}=6+r^{2}-4r+4$$

Add 6 and 4 to get 10.

$$r^{2}=10+r^{2}-4r$$

Subtract r^{2} from both sides.

$$r^{2}-r^{2}=10-4r$$

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

$$0=10-4r$$

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

$$10-4r=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$-4r=-10$$

Divide both sides by -4.

$$r=\frac{-10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out -2.

$$r=\frac{5}{2}$$