It is well known that there is no general formula for the solution of the quintic. Of course, what this really means is that there is no general formula that only involves addition, subtraction, multiplication, division, and the extraction of $n$-th roots. Indeed, if one is allowed to use the Bring radical, that is, solutions of the equation $x^5+x+a=0$, then it is indeed possible to solve any quintic. It would seem that if one introduced higher order Bring radicals, it would be possible to solve polynomials of higher degree. More precisely, define a Bring radical of order $n$ to be a continuous function $B\_n(t)$ such that $B\_n(t)$ is a solution to an $n$-th degree polynomial, one of whose coefficients is $t$. (Of course, I am being rather vague, most of these Bring radicals are only continuously definable on some proper subset of $\mathbb{C}$) It is trivial that any $n$-th degree polynomial may be solved by means of some $n$-th order Bring radical. However, it is not at all apparent that for some fixed $n$, there exist a finite collection $B\_n^1,B\_n^2,\cdots,B\_n^k$ such that *any* $n$-th degree polynomial may be solved using the $B\_n^i$. So my question is: 



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> Is it the case that for any $n$, there is a finite collection of Bring radicals that may be used to solve any $n$-th degree polynomial? 
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Another question, to which the answer is most likely negative, is whether there exists a finite collection $B\_{r\_1}^1,B\_{r\_2}^2,\cdots$ of Bring radicals such that any polynomial of any degree is solvable using the $B\_{r\_i}^j$.


**Edit**: My definition of a higher-order Bring radical was rather narrow. I'd also be interested in any answer that involved Bring radicals $B\_n(t)$ that were solutions to a polynomial of the form $x^n+p\_{n-1}(t)x^{n-1}+\cdots+p\_{1}(t)x+p\_{0}(t)$. The general idea concerns whether or not one can solve *all* $n$-th degree polynomials just be adjoining the roots of some finite family of them polynomials, whose coefficients depend smoothly on $t$.



Using higher-order Bring radicals to solve arbitrary polynomials
The answer to the second question is "no". For a family of polynomials $p\_t$ depending polynomially on a complex parameter, as in the polynomials satisfied by your $B\_r(t)$, define its *Galois group* to be the group of permutations of the roots you see by moving around the branch points. (Assume that the roots of $p\_t$ are distinct for generic values of $t$ to make this work well. This is the same as the Galois group of $\mathcal{C}(t)(\text{roots of }p\_t)$ over $\mathcal{C}(t)$.) Then there are families of polynomials exhibiting an arbitrary finite group as its Galois group. Any finite family of $B^j(t)$ will only exhibit finitely many groups $G\_j$, and any tower of roots of the $B^j$ will give only groups whose composition factors are among the $G\_j$. Since there are infinitely many finite simple groups, you cannot acheive all possible finite groups this way.


However, I suspect the question was stronger than you meant to ask, since even the original conjecture that all polynomials are solvable by radicals wouldn't fit in to the framework of your second question.


