Given a solvable Galois group, to find a formula for the roots? Rereading my old abstract algebra textbook, there's a mention that for every polynomial with a solvable Galois group, there's a formula for the roots comparable to the classical formulae for polynomials of order 2, 3, and 4, but it doesn't elaborate.  So I was wondering - let's say you know the Galois group, but only have a symbolic representation of the coefficients, how do you come up with such a formula?
 A: $\def\cK{\mathcal{K}}\def\cF{\mathcal{F}}\def\cL{\mathcal{L}}\def\QQ{\mathbb{Q}}$Here is an almost answer to this question: Given $G$ a solvable subgroup of $S_n$, I'll show how to give a formula for the roots of a generic degree $n$ polynomial whose Galois group is contained in $n$. However, this formula may involve division by $0$ in "unlucky" cases.
Throughout this answer, we are going to need to maintain a careful separation between formal polynomials and those polynomials evaluated at particular complex numbers. I'll use capital letters for the former and lower case for the latter. I'll use caligraphic font for fields of formal polynomials and Roman font for subfields of $\mathbb{C}$.
Let $\cL = \QQ(X_1, X_2, \ldots,X_n)$, so $S_n$ acts on $\cL$. Let $\cK$ be the fixed field of $S_n$, so $\cK$ is generated by the elementary symmetric function $E_1$, $E_2$, ..., $E_n$. Let $\cF$ be the intermediate field fixed by $G$. Let $F$ be a primitive element for $\cF$ over $\cK$, with minimal polynomial $P(z)$ over $\cK$. We may assume that $F$ is a polynomial in $\QQ[X_1, \ldots, X_n]$, not just a rational function.
Now, $\cL/\cF$ is Galois with Galois group $G$, which we assumed solvable, so it is contained in a radical extension, and there are radical formulas for the $X_j$ in terms of the generators  of $\cF$, which is to say, in terms of $E_1$, $E_2$, ..., $E_n$ and $F$.
Now, suppose we had a particular polynomial $q(z)$ in $\QQ[z]$ with splitting field $L$. Let $x_1$, $x_2$, ..., $x_n$ be the roots of $f$ in $L$ and, embedding $\mathrm{Gal}(L/k)$ into $S_n$ by the action on the $x_j$, suppose that $\mathrm{Gal}(L/k) \subseteq G$. Let $e_1$, $e_2$, ..., $e_n$ be the elementary polynomials in the $x$'s, so the $e$'s lie in $k$. The polynomial $F(X_1, \ldots, X_n)$ is $G$-invariant so the evaluation $F(x_1, \ldots, x_n)$ is fixed by $\mathrm{Gal}(L/k)$ and is thus in $k$. Let $f = F(x_1, \ldots, x_n)$.
Now, the coefficients of the minimal polynomial $P(z)$ lie in $\cK = \QQ(E_1, E_2, \ldots, E_n)$. We can evaluate them at $e_1$, $e_2$, ..., $e_n$ and obtain a polynomial $p(z) \in \QQ[z]$, which $f$ will be a root of. Note that the formula for the coefficients of $p[z]$ in term of $e_1$, $e_2$, ..., $e_n$ is a universal formula, depending only on the group $G$.
For any particular $q(z)$, we may thus compute $p(z)$ and use the rational root theorem to test if it has a rational root $f$. If it does, we may then plug $e_1$, $e_2$, ..., $e_n$, $f$ into the radical formulas for $x_j$ in terms of $E_1$, $E_2$, ..., $E_n$, $F$.
If this process does not involve dividing by $0$, and if the correct values of radicals are chosen, it will give a root of $q(z)$. The bold phrase is the rub. The primitive element $F$ only generates $\cF$ over $\cK$ as a field, so there may well be division in expressing the other elements of $\cF$ in terms of it. 
If I get a chance later, I'll write up some examples. Doing this in practice is pretty labor intensive.

Here is a worked example focusing on constructiblity rather than radicals. Recall that the roots of a polynomial are constructible if and only if the Galois group is a $2$-group. So the roots of a degree $n$ polynomial are constructible if and only if the Galois group is contained in a Sylow $2$-subgroup of $S_n$. Let's see how this works for $n=4$. An explicit Sylow is the dihedral group $G := \langle (12)(34), (1234) \rangle$. 
Let $\cL = \QQ(X_1, X_2, X_3, X_4)$, let $\cF$ be the subfield fixed by $G$ and let $\cK = \QQ(X_1, X_2, X_3, X_4)^{S_4} = \QQ(E_1, E_2, E_3, E_4)$. A primitive element for $\cF/\cK$ is $F:=X_1 X_3 + X_2 X_4$. The minimal polynomial of $F$ over $\cK$ is
$$(z-X_1 X_2 - X_3 X_4)(z-X_1 X_3 - X_2 X_4)(z-X_1 X_4 - X_2 X_3)$$
$$= z^3 - E_2 z^2 + (E_1 E_3 - E_4) z - (E_3^2 + E_1^2 E_4 - 4 E_2 E_4). \qquad(\ast)$$
Setting $B = X_1+X_3$ and $C = X_1 X_3$, we have
$$X_1^2 - B X_1 +  C= 0$$
$$B^2 - E_1 B + (E_2 - F)=0$$
$$C^2 - F C + E_4 = 0$$
and thus
$$X_1 = \frac{B + \sqrt{B^2-4C}}{2}\quad B = \frac{E_1+\sqrt{E_1^2 - 4 E_2 + 4 F}}{2} \quad C = \frac{F+\sqrt{F^2 - 4 E_4}}{2}. \qquad(\dagger)$$
(Remark: $X_1$ is fixed by $\langle (24) \rangle$, $B$ and $C$ are fixed by $\langle (13), (24) \rangle$ and $F$ is fixed by $G$. Each containment of groups is index $2$, so it makes sense that we take a single square root each time.)
Thus, to test whether a quartic polynomial $z^4 - e_1 z^3 + e_2 z^2 - e_3 z + e_4$ has a constructible root, first see plug in $e_j$ for $E_j$ to $(\ast)$ and see if the resulting equation has a rational root $f$. If so, plug $e_j$ for $E_j$ and $f$ for $F$ in $(\dagger)$ to get a formula for that root. (Note that all the $2$-Sylows of $S_4$ are conjugate so, if the Galois group is contained in a different $2$-Sylow, one of the other factors in $(\ast)$ will contribute a root and everything will proceed as before.)
I remember seeing a similar worked example somewhere for $S_5$ and its $20$ element solvable subgroup; I'll see if I can track it down.
