Solve 4+x^2=16+left(4-xright)^2

$$4+ { x }^{ 2 } =16+ { \left(4-x \right) }^{ 2 }$$

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps for Solving Linear Equation

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

$$4+x^{2}=16+16-8x+x^{2}$$

Add 16 and 16 to get 32.

$$4+x^{2}=32-8x+x^{2}$$

Add 8x to both sides.

$$4+x^{2}+8x=32+x^{2}$$

Subtract x^{2} from both sides.

$$4+x^{2}+8x-x^{2}=32$$

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

$$4+8x=32$$

Subtract 4 from both sides.

$$8x=32-4$$

Subtract 4 from 32 to get 28.

$$8x=28$$

Divide both sides by 8.

$$x=\frac{28}{8}$$

Reduce the fraction \frac{28}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{7}{2}$$