As the others have commented, to know when a quintic (or higher) is solvable in radicals requires Galois theory. However, there is a rather simple aspect when it is not solvable that is easily understood and can be used as a litmus test.
Theorem: An irreducible equation of prime degree $p>2$ that is solvable in radicals has either $1$ or $p$ real roots.
(Irreducible, simply put, means it has no rational roots.) By sheer coincidence, the irreducible quintic you chose has $3$ real roots so, by looking at its graph, you can indeed tell at a glance that this is not solvable in radicals. Going higher, if an irreducible septic has $3$ or $5$ real roots, then you automatically know it is not solvable. And so on.
P.S. And before you ask, it does not work the other direction: if it has $1$ or $p$ real roots, it does not imply it is solvable in radicals. It is a necessary but not sufficient condition.