Question 886344

find 3 consecutive odd integers such that are twice the product of the first two is 7 more than the product of the last two.
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Let first integer be F
Then 2nd is: F + 2, and 3rd is: F + 4
Therefore, 2(F)(F + 2) = (F + 2)(F + 4) + 7
{{{2F^2 + 4F = F^2 + 6F + 8 + 7}}}
{{{2F^2 - F^2 + 4F - 6F - 15 = 0}}}
Solve for F, the 1st integer, and then determine the other 2 integers. Note that 2 values will ensue for
the 1st integer, F, thus 2 values for the 2nd and 3rd integers.