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

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

Solve for a

$$a=-2$$

Steps for Solving Linear Equation

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

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

Subtract 5 from 5 to get 0.

$$16+8a+a^{2}+a^{2}=a^{2}+a^{2}$$

Combine a^{2} and a^{2} to get 2a^{2}.

$$16+8a+2a^{2}=a^{2}+a^{2}$$

Combine a^{2} and a^{2} to get 2a^{2}.

$$16+8a+2a^{2}=2a^{2}$$

Subtract 2a^{2} from both sides.

$$16+8a+2a^{2}-2a^{2}=0$$

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

$$16+8a=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$8a=-16$$

Divide both sides by 8.

$$a=\frac{-16}{8}$$

Divide -16 by 8 to get -2.

$$a=-2$$