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

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

Solve for x

$$x=3$$

Steps for Solving Linear Equation

Calculate 4 to the power of 2 and get 16.

$$16+x^{2}=\left(8-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

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

Add 16x to both sides.

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

Subtract x^{2} from both sides.

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

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

$$16+16x=64$$

Subtract 16 from both sides.

$$16x=64-16$$

Subtract 16 from 64 to get 48.

$$16x=48$$

Divide both sides by 16.

$$x=\frac{48}{16}$$

Divide 48 by 16 to get 3.

$$x=3$$