I2.1 Suppose $P$ is an integer and $5<P<20$. If the roots of the equation $x^{2}-2(2 P-3) x+4 P^{2}-14 P+8=0$ are integers, find the value of $P$.
$$
\begin{array}{l}
\Delta=4(2 P-3)^{2}-4\left(4 P^{2}-14 P+8\right)=m^{2} \\
\left(\frac{m}{2}\right)^{2}=4 P^{2}-12 P+9-4 P^{2}+14 P-8=2 P+1 \\
\because 5<P<20 \therefore 11<2 P+1<41
\end{array}
$$

The only odd square lying in this interval is 25
$$
\begin{array}{l}
\Rightarrow 2 P+1=25=5^{2} \\
\therefore P=12
\end{array}
$$