SOLUTION: Find three consecutive odd integers such that the product of the first two minus six times the third is 21.

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Question 629291: Find three consecutive odd integers such that the product of the first two minus six times the third is 21.
Answer by dfrazzetto(283) About Me  (Show Source):
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2n+1, 2n+3, 2n+5
(2n+1)(2n+3) - 6(2n+5) = 21
4n^2 + 8n + 3 - 12n - 30 = 21
4n^2 -4n -27 -21 = 0
4n^2 - 4n - 48 = 0
n^2 - n - 12 = 0
(n+3)(n-4) = 0
n = -3, 4
two sets of numbers satisfy:
{-5, -3, -1}
{9, 11, 13}