Solve x^2=(x-2)^2+8^2

$$x ^ { 2 } = ( x - 2 ) ^ { 2 } + 8 ^ { 2 }$$

Solve for x

$$x=17$$

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}.

$$x^{2}=x^{2}-4x+4+8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$x^{2}=x^{2}-4x+4+64$$

Add 4 and 64 to get 68.

$$x^{2}=x^{2}-4x+68$$

Subtract x^{2} from both sides.

$$x^{2}-x^{2}=-4x+68$$

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

$$0=-4x+68$$

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

$$-4x+68=0$$

Subtract 68 from both sides. Anything subtracted from zero gives its negation.

$$-4x=-68$$

Divide both sides by -4.

$$x=\frac{-68}{-4}$$

Divide -68 by -4 to get 17.

$$x=17$$