Solve x^2+64=left(12-xright)^2

$${ x }^{ 2 } +64= { \left(12-x \right) }^{ 2 }$$

Solve for x

$$x = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-x\right)^{2}.

$$x^{2}+64=144-24x+x^{2}$$

Add 24x to both sides.

$$x^{2}+64+24x=144+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+64+24x-x^{2}=144$$

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

$$64+24x=144$$

Subtract 64 from both sides.

$$24x=144-64$$

Subtract 64 from 144 to get 80.

$$24x=80$$

Divide both sides by 24.

$$x=\frac{80}{24}$$

Reduce the fraction \frac{80}{24} to lowest terms by extracting and canceling out 8.

$$x=\frac{10}{3}$$