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