Solve (x+10)^2=(5-x)^2

$$( x + 10 ) ^ { 2 } = ( 5 - x ) ^ { 2 }$$

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(x+10\right)^{2}.

$$x^{2}+20x+100=\left(5-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-x\right)^{2}.

$$x^{2}+20x+100=25-10x+x^{2}$$

Add 10x to both sides.

$$x^{2}+20x+100+10x=25+x^{2}$$

Combine 20x and 10x to get 30x.

$$x^{2}+30x+100=25+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+30x+100-x^{2}=25$$

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

$$30x+100=25$$

Subtract 100 from both sides.

$$30x=25-100$$

Subtract 100 from 25 to get -75.

$$30x=-75$$

Divide both sides by 30.

$$x=\frac{-75}{30}$$

Reduce the fraction \frac{-75}{30} to lowest terms by extracting and canceling out 15.

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