Solve 17^2-x^2=10^2-(21-x)^2

$$17 ^ { 2 } - x ^ { 2 } = 10 ^ { 2 } - ( 21 - x ) ^ { 2 }$$

Solve for x

$$x=15$$

Steps for Solving Linear Equation

Calculate 17 to the power of 2 and get 289.

$$289-x^{2}=10^{2}-\left(21-x\right)^{2}$$

Calculate 10 to the power of 2 and get 100.

$$289-x^{2}=100-\left(21-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(21-x\right)^{2}.

$$289-x^{2}=100-\left(441-42x+x^{2}\right)$$

To find the opposite of 441-42x+x^{2}, find the opposite of each term.

$$289-x^{2}=100-441+42x-x^{2}$$

Subtract 441 from 100 to get -341.

$$289-x^{2}=-341+42x-x^{2}$$

Subtract 42x from both sides.

$$289-x^{2}-42x=-341-x^{2}$$

Add x^{2} to both sides.

$$289-x^{2}-42x+x^{2}=-341$$

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

$$289-42x=-341$$

Subtract 289 from both sides.

$$-42x=-341-289$$

Subtract 289 from -341 to get -630.

$$-42x=-630$$

Divide both sides by -42.

$$x=\frac{-630}{-42}$$

Divide -630 by -42 to get 15.

$$x=15$$