Solve x^2=(x-4)^2+8^2

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

Solve for x

$$x=10$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}=x^{2}-8x+16+8^{2}$$

Calculate 8 to the power of 2 and get 64.

$$x^{2}=x^{2}-8x+16+64$$

Add 16 and 64 to get 80.

$$x^{2}=x^{2}-8x+80$$

Subtract x^{2} from both sides.

$$x^{2}-x^{2}=-8x+80$$

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

$$0=-8x+80$$

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

$$-8x+80=0$$

Subtract 80 from both sides. Anything subtracted from zero gives its negation.

$$-8x=-80$$

Divide both sides by -8.

$$x=\frac{-80}{-8}$$

Divide -80 by -8 to get 10.

$$x=10$$