Solve (x-4)^2+64=x^2

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

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}.

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

Add 16 and 64 to get 80.

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

Subtract x^{2} from both sides.

$$x^{2}-8x+80-x^{2}=0$$

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

$$-8x+80=0$$

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.

$$x=10$$