smc

By Vieta's, $r+s=-\frac{B}{A}$, $rs=\frac{C}{A}$, and $r^2+s^2=-p$. Note that $(r+s{)}^2=r^2+s^2+2rs$. Therefore, $(r+s{)}^2-2rs=r^2+s^2$, or $-{\left(\frac{B}{A}\right)}^2-\frac{2C}{A}=-p$. Simplifying, $\frac{B^2-2CA}{A^2}=-p$. Finally, multiply both sides by $-1$ to get $p=\frac{2CA-B^2}{A^2}$, making the answer $\boxed{\frac{2AC-B^2}{A^2}}$.