2 positive consecutive odd integers such that the square of the first plus 3 times the second = 24

[math_celebrity]

2 positive consecutive odd integers such that the square of the first plus 3 times the second = 24

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

Square of the first plus 3 times the second = 24:
n^2 + 3(n + 2) = 24
n^2 + 3n + 6 = 24

Subtract 24 from each side:
n^2 + 3n - 18 = 0

Using our quadratic equation calculator, we have:
n = (3, -6)

We're told the numbers are positive, so n = 3 is our first postive odd integer.
The next one is 3 + 2 = 5.

Our solution is (3, 5)