Given a group $G$ of order 440, it has a unique subgroup of order 11 which is normal in $G$. Let it be denoted $H$.
$H$ is clearly solvable, if $G/H$ was solvable, so it would be $G$. However I cannot seem to be able to show this last bit. Any idea on how to show that $G/H$ is solvable? Also, any other idea on how to prove that $G$ is solvable would be greatly appreciated.
$G/H$ is a group of order $40$, so your same Sylow technique shows that it has a unique normal subgroup of order $5$, and once again reduces to a smaller case: now we can assume $|G|=8$. This is a $p$-group for $p=2$, therefore it is nilpotent and solvable.