Solve 40^2+left(80-xright)^2=x^2

$${ 40 }^{ 2 } + { \left(80-x \right) }^{ 2 } = { x }^{ 2 }$$

Solve for x

$$x=50$$

Steps for Solving Linear Equation

Calculate 40 to the power of 2 and get 1600.

$$1600+\left(80-x\right)^{2}=x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(80-x\right)^{2}.

$$1600+6400-160x+x^{2}=x^{2}$$

Add 1600 and 6400 to get 8000.

$$8000-160x+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$8000-160x+x^{2}-x^{2}=0$$

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

$$8000-160x=0$$

Subtract 8000 from both sides. Anything subtracted from zero gives its negation.

$$-160x=-8000$$

Divide both sides by -160.

$$x=\frac{-8000}{-160}$$

Divide -8000 by -160 to get 50.

$$x=50$$