Two years ago, a man's age was three times the square of his son's age. In three years’ time, his age will be four times his son's age. Find their present ages.
Let son’s age 2 years ago be x years, Then
man’s age 2 years ago be 3x2 years
son’s present age = (x + 2) years
man’s present age = (3x2 + 2)years
In three years’ time :
son’s age = (x + 2 + 3) = (x + 5) years
man’s age = (3x2 + 2 + 3)years = (3x2 + 5) years
According to question
Man’s age = 4 son’s age
3x2 + 5 = 4(x + 5)
3x2 + 5 = 4x + 20
3x2 – 4x – 15 = 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 = 3 b = – 4 c = – 15
= 3. – 15 = – 45
And either of their sum or difference = b
= – 4
Thus the two terms are – 9 and 5
Difference = – 9 + 5 = – 4
Product = – 9.5 = – 45
3x2 – 9x + 5x – 15 = 0
3x(x – 3) + 5(x – 3) = 0
(x – 3) (3x + 5) = 0
(x – 3) = 0 or (3x + 5) = 0
x = 3 or x = – 5/3 (age cannot be negative)
x = 3
son’s present age = (3 + 2) = 5years
man’s present age = (3.32 + 2) = 29years