I got interested in this subject last year (2011) and just got round to writing up [some notes][1] which I hope may be of use. I also have a python script hosted [here][2], which implements Klein's icosahedral solution of the quintic, as well as a brief summary of what it does [here][3].

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:
  <font color="red">
  $$
    y = -\gamma\cdot\frac{f(z)}{H(z)/B(z)} -
        \frac{7\gamma^2 + 9\nabla}{2\gamma(\gamma^4 + 648)}
        \cdot\frac{D(z)T(z)}{f(z)^2H(z)/B(z)}
  $$
  </font>
  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.

  [1]: https://arxiv.org/abs/1308.0955
  [2]: https://github.com/ocfnash/icosahedral_quintic
  [3]: http://olivernash.org/2012/02/05/on-kleins-icosahedral-solution-of-the-quintic/index.html