Question

How do you equate this?
9^2+x^2=4^2+(10-x)^2

Please explain clearly that anyone could understand it easily.
(Guaranteed medal for a good answer)

Answer 1

Combine the x^2 together on one side and the constants on another side.
Set them equal to each other.

x^2 - (10-x)^2 = 4^2 - 9^2

Answer 2

Expanding it, you get x^2 - 100 +20x - x^2 = 4^2 - 9^2
The x^2 , cancels out. You are left with
-100 + 20x = 4^2 - 9^2
20x = 4^2 - 9^2 + 100 = 35
x = 35/20

Answer 3

Thanks!

Answer 4

9^2+x^2=4^2+(10-x)^2
Sort out the constants first, so 9^2 = 81 and 4^2 = 16
Rewrite
81 + x^2 = 16 +(10-x)^2
Expand (10-x)^2 which gives 100 -20x + x^2
Rewrite
81 + x^2 = 16 +100 -20x + x^2 = 116 -20x + x^2
Subtract x^2 from both sides
81 = 116 -20x
Subtract 116 from both sides
-35 = -20x
divide both sides by -5
7 = 4x
Divide both sides by 4
7/4 = x