Solve (x-4)^2=(x+5)^2

$$( x - 4 ) ^ { 2 } = ( x + 5 ) ^ { 2 }$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16=\left(x+5\right)^{2}$$

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

$$x^{2}-8x+16=x^{2}+10x+25$$

Subtract x^{2} from both sides.

$$x^{2}-8x+16-x^{2}=10x+25$$

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

$$-8x+16=10x+25$$

Subtract 10x from both sides.

$$-8x+16-10x=25$$

Combine -8x and -10x to get -18x.

$$-18x+16=25$$

Subtract 16 from both sides.

$$-18x=25-16$$

Subtract 16 from 25 to get 9.

$$-18x=9$$

Divide both sides by -18.

$$x=\frac{9}{-18}$$

Reduce the fraction \frac{9}{-18} to lowest terms by extracting and canceling out 9.

$$x=-\frac{1}{2}$$