Find three positive consecutive integers such that the product of the first and the second is two more than three times the third

3 consecutive integers: a+2=b+1=c

a*b+2>3*c

substitute a and b definitions with c:
(c-2)*(c-1)+2>3c
c^2-c-2c+2+2>3c
c^2-3c+4>3c
c^2-6c+4>0
factoring
c^2-6c+4=(c-3)^2-5

(c-3)^2-5>0
(c-3)^2>5
c-3>sqrt(5)
c>sqrt(5)+3

c is an integer so we round up and don't have to calculate it exactly:
c=6, because they are consecutive a=4, b=5
4*5+2>3*6
20+2>18