Solve 4+left(6-xright)^2=x^2

$$4+ { \left(6-x \right) }^{ 2 } = { x }^{ 2 }$$

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(6-x\right)^{2}.

$$4+36-12x+x^{2}=x^{2}$$

Add 4 and 36 to get 40.

$$40-12x+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$40-12x+x^{2}-x^{2}=0$$

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

$$40-12x=0$$

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

$$-12x=-40$$

Divide both sides by -12.

$$x=\frac{-40}{-12}$$

Reduce the fraction \frac{-40}{-12} to lowest terms by extracting and canceling out -4.

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