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

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

Solve for x

$$x=6$$

Steps for Solving Linear Equation

Calculate 8 to the power of 2 and get 64.

$$x^{2}+64=\left(x+4\right)^{2}$$

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

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

Subtract x^{2} from both sides.

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

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

$$64=8x+16$$

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

$$8x+16=64$$

Subtract 16 from both sides.

$$8x=64-16$$

Subtract 16 from 64 to get 48.

$$8x=48$$

Divide both sides by 8.

$$x=\frac{48}{8}$$

Divide 48 by 8 to get 6.

$$x=6$$