Question 320442
Let one integer be x
Then the second integer = 21 - x.

Their product = 108
ie. x(21-x) = 108
21x - x^2 = 108
21x - x^2 - 108 = 0
-x^2 + 21x - 108 = 0
Multiplying the whole with -1, (in order to make the co-efficient of x^2 +ve)

x^2 - 21x + 108 = 0 -------- (1)

[Now we have to find two numbers such that their product is 108 and sum is -21. and these two numbers will eb factors of 108. They are -9 and -12]

So (1) can be written as
x^2 - 9x - 12x + 108 = 0
x(x-9) - 12(x-9) = 0
(x-9) (x-12) = 0
So either x-9 = 0   or x-12 = 0
           x = 9    or  x = 12
These are the required integers