Solve x^2+1=left(3-xright)^2

$${ x }^{ 2 } +1= { \left(3-x \right) }^{ 2 }$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$x^{2}+1=9-6x+x^{2}$$

Add 6x to both sides.

$$x^{2}+1+6x=9+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+1+6x-x^{2}=9$$

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

$$1+6x=9$$

Subtract 1 from both sides.

$$6x=9-1$$

Subtract 1 from 9 to get 8.

$$6x=8$$

Divide both sides by 6.

$$x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$