The most generic radicals-solvable quintic

Question

It's well known that it is impossible to solve a generic quintic equation in terms of radicals involving its coefficients. However: what's the "most generic" quintic equation that is still possible to solve using radicals?

If we think of a quintic as $ax^5+bx^4+cx^3+dx^2+ex+f = 0$ then "the most generic" here would mean such an equation that describes the largest solvable subset of quintics. There certainly exist some subsets of quintics that are solvable, a good example is $ax^5+b=0$ , but I want to know what's the most generic look of a solvable quintic is.

A fascinating book: http://people.reed.edu/~jerry/Quintic/quintic.pdf — Jean Marie, Feb 12 '22 at 13:43
I'm now curious if the set of solvable quintics is connected (viewed as a subset of R^6), simply connected, etc. — Beren Gunsolus, Feb 12 '22 at 13:56
Have you seen this question and its answers https://math.stackexchange.com/q/1019449/305862 ? — Jean Marie, Feb 15 '22 at 08:42
Would something like this post help? Also this paper if a polynomial can be depressed into a trinomial. — Тyma Gaidash, Feb 19 '22 at 15:56
@BerenGunsolus This question doesn't make sense: One can only ask whether a quintic is solvable when it has rational coefficients. — David E Speyer, Feb 23 '22 at 14:53
Suppose I mean for it to be solvable with radicals, allowing use of the coefficients. In this sense x^5=e is "solvable" — Beren Gunsolus, Feb 23 '22 at 23:16
The answer is if one of the roots is made from radicals, then so must the other $4$ since factoring the general quintic using a radical root will give a general quartic which always has radical roots. Right? — Тyma Gaidash, Feb 24 '22 at 16:55

The most generic radicals-solvable quintic

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