Solve (3+x)^2=(3-x)^2+16

$$( 3 + x ) ^ { 2 } = ( 3 - x ) ^ { 2 } + 16$$

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}.

$$9+6x+x^{2}=\left(3-x\right)^{2}+16$$

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

$$9+6x+x^{2}=9-6x+x^{2}+16$$

Add 9 and 16 to get 25.

$$9+6x+x^{2}=25-6x+x^{2}$$

Add 6x to both sides.

$$9+6x+x^{2}+6x=25+x^{2}$$

Combine 6x and 6x to get 12x.

$$9+12x+x^{2}=25+x^{2}$$

Subtract x^{2} from both sides.

$$9+12x+x^{2}-x^{2}=25$$

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

$$9+12x=25$$

Subtract 9 from both sides.

$$12x=25-9$$

Subtract 9 from 25 to get 16.

$$12x=16$$

Divide both sides by 12.

$$x=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

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