To solve the general quintic using elliptic functions, one way is to reduce it to *Bring-Jerrard form*,

$$x^5-x+ d = 0\tag{1}$$

a transformation which can be done in radicals. (See this post.) To solve $(1)$, define,

$$k = \tan\left(\tfrac{1}{4}\arcsin\Big(\frac{16}{25\sqrt{5}\,d^2}\Big)\right)\tag{2}$$

$$p = i\frac{K(k')}{K(k)} = i\,\frac{\text{EllipticK[1-m]}}{\text{EllipticK[m]}}\tag3$$

with the *complete elliptic integral of the first kind* $K(k)$ and *elliptic parameter* $m=k^2$ (with $p$ also given in *Mathematica* syntax above).

**Method 1**: For $j=0,1,2,3,4$, let,

$$S_j={u}^j \frac{\sqrt{2}\,\eta(\tau_j)\,\eta^2(4\,\tau_j)}{\eta^3(2\,\tau_j)}$$

$$\tau_j = \tfrac{1}{10}(p+2j)$$

$$u=\zeta_8 = \exp(2 \pi i/8)$$

$$S_5=\frac{\sqrt{2}\,\eta(\frac{5p}{2})\,\eta^2(10p)}{\eta^3(5p)}$$

where $\eta(\tau)$ is the *Dedekind eta function*, then,

$$x = \frac{\pm 1}{2\cdot 5^{3/4}}\frac{(k^2)^{1/8}}{\sqrt{k(1-k^2)}}(S_0+S_5)(S_1+i\,S_4)(i\,S_2+S_3) \tag{4}$$

with the sign chosen appropriately.

$\color{blue}{\text{Remark}}$: The five roots $x_n$can be found by using $p_n = i\frac{K(k')}{K(k)}+16n$ for $n = 0,1,2,3,4$.

**Method 2**: For $j=0,1,2,3,4$, let,

$$T_j =\left(\frac{\vartheta_2(0,w^j q^{1/5})}{\vartheta_3(0,w^j q^{1/5})}\right)^{1/2}$$

$$q = \exp(i \pi p_0)$$

$$w = \zeta_5 = \exp(2 \pi i/5)$$

$$T_5 =\frac{q^{5/8}}{(q^5)^{1/8}}\left(\frac{\vartheta_2(0,q^{5})}{\vartheta_3(0, q^{5})}\right)^{1/2}$$

with $\vartheta_n(0,q)$ as the *Jacobi theta functions*, then,

$$x = \frac{\pm{\zeta_8}}{2\cdot 5^{3/4}}\frac{(k^2)^{1/8}}{\sqrt{k(1-k^2)}}(T_0+T_5)(T_1-i\,T_4)(T_2+T_3) \tag{5}$$

and the sign of $\zeta_8$ chosen appropriately. Note that $(S_j)^8 = (T_j)^8$.

$\color{blue}{\text{Remark}}$: One can also find the other roots $x_i$, but is not as simple as in Method 1. (As one can see, you need *much* more than radicals to solve the general quintic.)

*Example*. Let,

$$x^5-x+1 = 0\tag6$$

so $d = \pm1$. Plugging it into $(2)$ gives $k \approx 0.072696$, and $m=k^2$, so $p \approx 2.550572\,i$. Since both methods use a square root, using the formulas one eventually finds,

$$x = \mp\,1.1673039\dots$$

$\color{blue}{\text{Afterword}}$ (Added June 2015): Given the *nome* $q = \exp(i \pi \tau)$, the two methods involve the function known either as the *modular lambda function* or *elliptic lambda function*, $\lambda(\tau)$ which has a beautiful q-continued fraction,

$$\big(\lambda(\tau)\big)^{1/8} = \frac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\,\eta^2(2\tau)}{\eta^3(\tau)} = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^{1/2} = \cfrac{\sqrt{2}\,q^{1/8}}{1+\cfrac{q}{1+q+\cfrac{q^2}{1+q^2+\cfrac{q^3}{1+q^3+\ddots}}}}$$

studied by Ramanujan (who also had his own method to solve solvable quintics).