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

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

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}.

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

Add 16 and 4 to get 20.

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

Subtract x^{2} from both sides.

$$20-8x+x^{2}-x^{2}=0$$

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

$$20-8x=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$-8x=-20$$

Divide both sides by -8.

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

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

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