Is there something like Cardano's method for a SOLVABLE quintic.
Yes, there is a general expression for solutions of solvable quintics. A published account, including the relevant formulas, appears in
David S. Dummit. Solving solvable quintics, Math. Comp., 57 (195), (1991), 387–401. MR1079014 (91j:12005). Corrigenda, Math. Comp., 59 (199), (1992), 309. MR1166516.
The corrigenda is actually a microfiche supplement containing an appendix to the paper, that was inadvertently not published with the paper itself. It contains some of the formulas needed for the explicit computation of the roots. To give you an idea of the complexity of the task, the appendix runs for $43$ pages.
From the paper:
It is well known that an irreducible quintic with coefficients in the rational numbers $\mathbb Q$ is solvable by radicals if and only if its Galois group is contained in the Frobenius group $F_{20}$ of order $20$, i.e., if and only if the Galois group is isomorphic to $F_{20}$ , to the dihedral group $D_{10}$ of order $10$, or to the cyclic group $\mathbb Z/5\mathbb Z$.
Explicitly associated to an irreducible quintic $f$ there is a resolvent sextic that Dummit calls $f_{20}$, and $f$ is solvable by radicals iff $f_{20}$ has a rational root, in which case $f_{20}$ factors as a linear term times an irreducible quintic. When this happens, explicit criteria are given for which of the three possibilities above actually occurs, together with the relevant formulas. The paper includes examples showing all three cases are possible.
Marc is correct, the Galois group for a solvable irreducible quintic can only have order $5,10$ or $20,$ since these are the only orders of solvable subgroups of $S_5$ which have order divisible by $5,$ so in some ways the possible structures for the Galois group of a solvable quintic are less messy than the structure of the symmetric group $S_4.$ Following the arguments in Galois Theory texts for the relevant Galois groups should be feasible( and is probably already done explicitly somewhere). It might help to note that such an irreducible quintic has either $1$ or $5$ real roots.