Answer
The numbers are $8\ \text{and}\ 12$.
Work Step by Step
Let us assume $ x $ and $ y $ be the two numbers such that:
$\begin{align}
& x+y=20\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& xy=96\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
From equation (I):
$ y=20-x $ (III)
Substitute $ y=20-x $ in equation (II):
$\begin{align}
& x\left( 20-x \right)=96 \\
& 20x-{{x}^{2}}=96 \\
& {{x}^{2}}-8x+96=0 \\
& \left( x-8 \right)\left( x-12 \right)=0
\end{align}$
Now,
$ x=8,12$
Substitute $ x=8$ in equation (III):
$\begin{align}
& y=20-2 \\
& y=12 \\
\end{align}$
Substitute $x$ in equation (III)
Thus, the numbers are 8 and 12.
