So first, consider nth roots. These correspond to particularly simple field extensions. If I take two choices of nth root, their ratio is always an nth root of

So first, consider nth roots. These correspond to particularly simple field extensions. If I take two choices of nth root, their ratio is always an nth root of unity. This more or less forces the symmetry group to be a subgroup of the nth roots of unity, i.e. Z/n. Now if you have a general expression built out of extracting nth roots for various n, that might generate some big complicated field extension. But it corresponds to a tower of field extensions which are all just extracting nth roots. Thus the intermediate symmetry groups are still all subsets of Z/n for various n. The Galois correspondence tells you that the symmetry group of the entire extension, call it G, has a corresponding decomposition: there is an increasing sequence of subgroups 1 < G_1 < … < G_k = G such that each intermediate quotient group G_{i+1}/G_i is a subgroup of Z/n for some n. Groups with such a decomposition are called solvable. Insolvability of the quintic then boils down to this: the symmetry group of any degree k polynomial must be a subgroup of S_k. For k <= 4, S_k is solvable; for k >= 5, it is not. This last sentence is the closest thing I can offer to intuition. It is a very finite and tangible fact that one can write down and computationally verify for themselves. Like, you can write down the corresponding solvable decomposition of S_4, and you can see that S_5 is not solvable by verifying that A_5 is simple which itself boils down to a small number of cases.