Solve 1^2+(2-x)^2=(1+x)^2

$$1 ^ { 2 } + ( 2 - x ) ^ { 2 } = ( 1 + x ) ^ { 2 }$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

Steps for Solving Linear Equation

Calculate 1 to the power of 2 and get 1.

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

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

$$1+4-4x+x^{2}=\left(1+x\right)^{2}$$

Add 1 and 4 to get 5.

$$5-4x+x^{2}=\left(1+x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+x\right)^{2}.

$$5-4x+x^{2}=1+2x+x^{2}$$

Subtract 2x from both sides.

$$5-4x+x^{2}-2x=1+x^{2}$$

Combine -4x and -2x to get -6x.

$$5-6x+x^{2}=1+x^{2}$$

Subtract x^{2} from both sides.

$$5-6x+x^{2}-x^{2}=1$$

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

$$5-6x=1$$

Subtract 5 from both sides.

$$-6x=1-5$$

Subtract 5 from 1 to get -4.

$$-6x=-4$$

Divide both sides by -6.

$$x=\frac{-4}{-6}$$

Reduce the fraction \frac{-4}{-6} to lowest terms by extracting and canceling out -2.

$$x=\frac{2}{3}$$