Ans:
Let age of first friend $=x$ years and let age of second friend $=(20-x)$ years Four years ago, age of first friend $=(x-4)$ years
Four years ago, age of second friend $=(20-x)-4=(16-x)$ years
According to given condition,
$$
\begin{array}{l}
(x-4)(16-x)=48 \\
\Rightarrow 16 x-x^{2}-64+4 x=48 \\
\Rightarrow 20 x-x^{2}-112=0 \\
\Rightarrow x^{2}-20 x+112=0
\end{array}
$$
Comparing equation, $x^{2}-20 x+112=0$ with general quadratic equation
$a x^{2}+b x+c=0$, we get $a=1, b=-20$ and $c=112$
Discriminant $=\mathrm{b}^{2}-4 \mathrm{ac}=(-20)^{2}-4(1)(112)=400-448=-48