Two years ago, a man's age was three times the square of his daughter's age . Three years hence , his age will be four times his daughter's age . Find their present ages.

Let present age of daughter be x years

Two years before daughter's age = (x - 2) years

Man's age two years before = 3(x - 2)²

So, present age of man = 3(x - 2)² + 2

Three year's later

Daughter's age = (x + 3) years

Man's age = (3(x - 2)² + 2 + 3) years

According to question,

$\Rightarrow (3(x - 2)^2 + 2 + 3) = 4(x + 3) \\[1em] \Rightarrow 3(x^2 + 4 - 4x) + 5 = 4x + 12 \\[1em] \Rightarrow 3x^2 + 12 - 12x + 5 = 4x + 12 \\[1em] \Rightarrow 3x^2 - 12x - 4x + 17 - 12 = 0 \\[1em] \Rightarrow 3x^2 - 16x + 5 = 0 \\[1em] \Rightarrow 3x^2 - 15x - x + 5 = 0 \\[1em] \Rightarrow 3x(x - 5) - 1(x - 5) = 0 \\[1em] \Rightarrow (3x - 1)(x - 5) = 0 \\[1em] \Rightarrow 3x - 1 = 0 \text{ or } x - 5 = 0 \\[1em] x = \dfrac{1}{3} \text{ or } x = 5$

Since, age cannot be in fraction so x ≠ $\dfrac{1}{3}$

∴ x = 5, 3(x - 2)² + 2 = 3(5 - 2)² + 2 = 3(3)² + 2 = 29

The present age of daughter is 5 years and of man is 29 years.