I got interested in this subject last year (2011) and just got round to writing up some notessome notes which I hope may be of use.
I also have a python script hosted here, which implements the Klein's icosahedral solution of the quintic linked from, as well as a brief summary of what it does this pagehere.
The geometry is easy to summarise: using a radical transformation, a quintic can be put in the form $y^5 + 5\alpha y^2 + 5\beta y + \gamma = 0$. The vector of ordered roots of such a quintic lies on the quadric surface $\sum y_i = \sum y_i^2 = 0$ in $\mathbb{P}^4$ and the reduced Galois group $A_5$ acts on the two families of lines in this doubly-ruled surface by permuting coordinates. The $A_5$ actions on these families, parameterized by $\mathbb{P}^1$, are equivalent to the action of the group of rotations of an icosahedron on its circumsphere and the quintic thus defines a point in the quotients — the icosahedral invariants of a quintic. Inverting either of these quotients (e.g., using the hypergeometric functions given below) is sufficient to allow us solve quintic (in rational functions).
Here's how it looks for a quintic in the simpler form: $$ y^5 + 5y + \gamma = 0 $$ (In fact any quintic can be put in this form using only radical transformations.)
Given such a quintic, set: $$ \nabla = \sqrt{\gamma^4 + 256}\\ Z = \frac{1}{2\cdot 1728}[2\cdot 1728 + 207\gamma^4 + \gamma^8 - \gamma^2 (81 + \gamma^4)\nabla]\\ z = \frac{{}_2F_1(\frac{31}{60}, \frac{11}{60}; \frac{6}{5}; Z^{-1})} {(1728Z)^{1/5}{}_2F_1(\frac{19}{60}, -\frac{1}{60}; \frac{4}{5}; Z^{-1})} $$ and: $$ f(z) = z(z^{10} + 11z^5 - 1)\\ H(z) = -(z^{20} + 1) + 228(z^{15} - z^5) - 494z^{10}\\ T(z) = (z^{30} + 1) + 522(z^{25} - z^5) - 10005(z^{20} + z^{10})\\ B(z) = -1 - z - 7(z^2 - z^3 + z^5 + z^6) + z^7 - z^8\\ D(z) = -1 + 2z + 5z^2 + 5z^4 - 2z^5 - z^6 $$ Then: $$ y = -\gamma\cdot\frac{f(z)}{H(z)/B(z)} - \frac{7\gamma^2 + 9\nabla}{2\gamma(\gamma^4 + 648)} \cdot\frac{D(z)T(z)}{f(z)^2H(z)/B(z)} $$ is a root.
Replacing $z$ with $e^{2\pi\nu i/5}z$ for $\nu=1, 2, 3, 4$ provides all the other roots.
Even in this rather gross explicit form, the link with regular solids is visible:
The roots of $f, H, T$ are, respectively, the locations of the projection of the vertices, face centres, and edge midpoints of a regular icosahedron onto its circumsphere (once this circumsphere has been identified with the extended complex plane by stereographic projection).
The roots of the last two polynomials, $B, D$ are, respectively, the locations of the vertices and face centres of a regular cube inscribed in the icosahedron.