2 consecutive odd integers such that their product is 15 more than 3 times their sum

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2 consecutive odd integers such that their product is 15 more than 3 times their sum.

Let the first integer be n. The next odd, consecutive integer is n + 2.

We are given the product is 15 more than 3 times their sum:
n(n + 2) = 3(n + n + 2) + 15

Simplify each side:
n^2 + 2n = 6n + 6 + 15
n^2 + 2n = 6n + 21

Subtract 6n from each side:
n^2 - 4n - 21 = 0

Type this problem into our search engine, and we get:
n = (-3, 7)

If we use -3, then the next consecutive odd integer is -3 + 2 = -1. So we have (-3, -1)
If we use 7, then the next consecutive odd integer is 7 + 2 = 9. So we have (7, 9)