Two natural numbers differ by 3 and their product is 504. Find the numbers.
Let the two natural numbers be x and (x + 3)
According to given condition,
x(x + 3) = 504
x2 + 3x – 504 = 0
Using the splitting middle term – the middle term of the general equation 
is divided in two such values that:
Product = a.c
For the given equation a = 1 b = 3 c = – 504
= 1. – 504 = – 504
And either of their sum or difference = b
= 3
Thus the two terms are 24 and – 21
Difference = 24 – 21 = 3
Product = 24. – 21 = – 504
x2 + 3x – 504 = 0
x2 + 24x – 21x – 504 = 0
x (x + 24) – 21(x + 24) = 0
(x + 24) (x – 21) = 0
(x + 24) = 0 or (x – 21) = 0
x = – 24 or x = 21
Case I: x = 21
x + 3 = 21 + 3 = 24
The numbers are (21, 24)
Case I: x = – 24
x + 3 = – 24 + 3 = – 21
The numbers are ( – 24, – 21)