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The age of a man is twice the square of the age of his son. Eight years hence, the age of the man will be $4$ years more than three times the age of his son. Find their present ages.
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Let present age of son = $x$ years
and present age of man = $2x^2$ years
A.T.Q.
$$\begin{aligned}& 3(x + 8) + 4 = 2x^2 + 8 \\ & \Rightarrow 2x^2 - 3x - 20 = 0 \\ & \Rightarrow (2x+5) (x - 4) = 0 \\ & x \neq -\frac{5}{2}\end{aligned}$$ So, $x = 4$
Present age of son = $4$ years
Present age of man = $32$ years
and present age of man = $2x^2$ years
A.T.Q.
$$\begin{aligned}& 3(x + 8) + 4 = 2x^2 + 8 \\ & \Rightarrow 2x^2 - 3x - 20 = 0 \\ & \Rightarrow (2x+5) (x - 4) = 0 \\ & x \neq -\frac{5}{2}\end{aligned}$$ So, $x = 4$
Present age of son = $4$ years
Present age of man = $32$ years