2. As shown in Figure 1, PB is a secant of circle $\odot O$ with radius 5, and the lengths of $PA$ and $PB$ are the two roots of the equation $x^{2}-10 x+16=0$ $(PA<PB)$, $PC$ is a tangent to $\odot O$, and $C$ is the point of tangency. Then the area of quadrilateral $PAOC$ is $\qquad$
2.14.

As shown in Figure 5, connect $O P$, and draw $O D \perp A B$, with $D$ being the foot of the perpendicular. Solving the equation $x^{2}-10 x+16=0$, we get $P A=2, P B=8$. Thus, $A D=3$, $O D=4$, and $P C=\sqrt{P A \cdot P B}=$
4. Therefore,
$S_{\text {quadrilateral PAOC }}$
$$
\begin{array}{l}
=S_{\triangle P O C}+S_{\triangle P D O}-S_{\triangle A D O} \\
=\frac{1}{2} \times 4 \times 5+\frac{1}{2} \times 4 \times 5-\frac{1}{2} \times 3 \times 4=14 .
\end{array}
$$