Find three consecutive odd integers such that 3 times the sum of all three is 24 more than the product if the first and second integers

3 consecutive odd integers : x, x + 2, x + 4

3 (x + x + 2 + x + 4) = x (x + 2) + 24

3 (3x + 6) = x^2 + 2x + 24

9x + 18 = x^2 + 2x + 24

x^2 - 7x + 6 = 0

(x - 1) (x - 6) = 0

x - 1 = 0

x = 1

x - 6 = 0

x = 6 (this will not work ... it is even)

ur 3 consecutive odd integers:

x = 1

x + 2 = 1 + 2 = 3

x + 4 = 1 + 4 = 5