Find all two digit positive integers in which the difference between the integer and the product of its two digits is 12

Let ab be a 2 digit number, then ab=10a+b, where a and b are digits, a≠0.

"the difference between the integer and the product of its two digits is 12"

10a+b - (a*b)=12

10a+b-ab=12

factorize b:

10a+b(1-a)=12

subtract 10 from both sides, to produce a factor (1-a) or (a-1) and then factorize:

10a-10+b(1-a)=2

10(a-1)+b(1-a)=2

(a-1)(10-b)=2

(a-1)(10-b) can be (1,2), (2, 1), (-1, -2), (-2, -1)

if a-1=1 and 10-b=2, then a=2, b=8
if a-1=2, and 10-b=1, then a=3, b=9
if a-1=-1, then a=0, which is not possible
if a-1=-2, then a=-1, which is not possible

so the solutions (a, b) are (2, 8) and (3, 9)

Then the integers are 28, 39

28, 39

The answer is a hope i