Question 115286
Let x and x + 2 be two consecutive odd integers. 
Their product is x (x+2)= x^2 + 2x. 
Their sum is x + x + 2 = 2x + 2.
Three times their sum = 3 (2x + 2) = 6x + 6.
Given that, their product is 15 more than three times their sum.
This can be mathematically written as x^2 + 2x = 6x + 6 + 15.
Simplifying, we get x^2 - 4x -21 = 0.

Solve. 

x^2 - 7x + 3x - 21 = 0
Or, x(x - 7) + 3(x - 7) = 0
Or, (x - 7) (x + 3) = 0
The solutions are x = 7, -3.

So, -3, -1 is a set of solution and 7,9 is another set of solution.