Three consecutive neg integer question

Question

Find three consecutive negative integers such that 3 times the product of the first and third exceeds twice the product of the first and second by 12.

Answer 1

Let the first negative number (least of three consecutive) be a-1; the second is a, the third is a+1 (it's closest to zero).

 

The problem says 3(a-1)(a+1) = 2(a-1)(a) + 12

 

Solve this  and keep only the value of a that is negative.