Question 1037647
Find three consecutive odd integers such that the square of the third integer plus the product of the other two integers is 268.
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Use n-2, n & n+2
(n+2)^2 + n(n-2) = 268
n^2+4n+4 + n^2-2n = 268
2n^2 + 2n + 4 = 268
n^2 + n - 132 = 0
n = -12, n = 11