smc

We are given that $x^2-5x+6<0$, which, when factored, gives $(x-2)(x-3)<0$. This has a solution of $2<x<3$, because the original quadratic is $\cup$-shaped, and thus dips below the x-axis between the roots. Since $x^2+5x+6$ has a vertex minimum at $x=-\frac{5}{2}$, so it is increasing on the interval $[2,3]$. Thus, evaluating $P$ at $x=2$ and $x=3$ will give our bounds, and doing so gives $20<P<30$, or $\boxed{20<P<30}$.