Let the required number be $x$ and $(28-x)$.

According to the given condition,

$\begin{array}{l}
x(28-x)=192 \\
\Rightarrow 28 x-x^{2}=192 \\
\Rightarrow x^{2}-28 x+192=0 \\
\Rightarrow x^{2}-16 x-12 x+192=0 \\
\Rightarrow x(x-16)-12(x-16)=0 \\
\Rightarrow(x-12)(x-16)=0 \\
\Rightarrow x-12=0 \text { or } x-16=0 \\
\Rightarrow x=12 \text { or } x=16
\end{array}$

When $x=12$

$28-x=28-12=16$

$\begin{array}{l}
\text { When } x=16 \\
28-x=28-16=12
\end{array}$

Hence, the required numbers are 12 and 16 .