Math Question

Let the two-digit number be 10A + B
The sum of digits of a two-digit number is 8 :
A + B = 8...(1)
If the digits are reserved, the new number is greater than the product of the digit by 38 :
10B + A = AB + 38...(2)
By (1) : A = 8 - B , sub it to (2) :
10B + (8 - B) = B(8 - B) + 38
9B + 8 = 8B - B^2 + 38
B^2 + B - 30 = 0
(B - 5)(B + 6) = 0
B = 5 or B = - 6 (rejected)
A = 8 - 5 = 3
The original number is 35