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

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

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

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}.

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

Add 16 and 4 to get 20.

$$x^{2}=20-8x+x^{2}$$

Add 8x to both sides.

$$x^{2}+8x=20+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+8x-x^{2}=20$$

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

$$8x=20$$

Divide both sides by 8.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$