Question 330195
Find three consecutive multiples of 5 such that the sum of the squares of the first two is 125 greater than the square of the third.

This is what I did, but it's wrong.

{{{(5N)^2 + (5N+5)^2 - 125 = (5N+10)^2}}}
This is ok, but you didn't square any terms.
{{{25n^2 + 25n^2 + 50n + 25 - 125 = 25n^2 + 100n + 100}}}
{{{50n^2 + 50n -100 = 25n^2 + 100n + 100}}}
{{{25n^2 - 50n - 200 = 0}}}
{{{n^2 - 2n - 8 = 0}}}
(n-4)*(n+2) = 0
n = 4
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25N + 25N + 25 - 125 = 25N + 100
25N - 100 = 100
25N = 200 
N = 8