Solve 40^2+(x-10)^2=x^2

$$40 ^ { 2 } + ( x - 10 ) ^ { 2 } = x ^ { 2 }$$

Solve for x

$$x=85$$

Steps for Solving Linear Equation

Calculate 40 to the power of 2 and get 1600.

$$1600+\left(x-10\right)^{2}=x^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-10\right)^{2}.

$$1600+x^{2}-20x+100=x^{2}$$

Add 1600 and 100 to get 1700.

$$1700+x^{2}-20x=x^{2}$$

Subtract x^{2} from both sides.

$$1700+x^{2}-20x-x^{2}=0$$

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

$$1700-20x=0$$

Subtract 1700 from both sides. Anything subtracted from zero gives its negation.

$$-20x=-1700$$

Divide both sides by -20.

$$x=\frac{-1700}{-20}$$

Divide -1700 by -20 to get 85.

$$x=85$$