(*This summarizes the accepted answer of Somos, and uses the j-function only*.) 

**Updated: Jan 16, 2023**

I just realized that since Somos' seven roots $r_k$ has an eta function $\eta(\tau)$ as a denominator, this can be incorporated into Ramanujan's parameterization for $\lambda, \mu, \nu$ so they are *theta quotients* and clearly ***radicals*** for $\tau=\sqrt{-d}$. Re-define,

$$a = \frac{-q^{25/56}f(-q,-q^6)}{\quad\eta(\tau)} = \frac{-q^{17/42}f(-q,-q^6)}{\quad f(-q)}$$
$$b = \frac{q^{9/56}f(-q^2,-q^5)}{\quad\eta(\tau)} =\; \frac{q^{5/42}f(-q^2,-q^5)}{\quad f(-q)}$$
$$c = \frac{q^{1/56}f(-q^3,-q^4)}{\quad\eta(\tau)} = \frac{q^{-1/42}f(-q^3,-q^4)}{\quad f(-q)}$$

which also satisfy,

$$a^3b+b^3c+c^3a = 0$$

Then *Klein's septic resolvent* reduces to the elegant formula for the [*j-function*][1] $j(\tau)$,

$$z\Big(z^3-\frac{8}{h^3}\sqrt{-7}\Big)\Big(z^3-\sqrt{-7}\Big)=\sqrt[3]{j(\tau)}$$

where $h=\frac{-1+\sqrt{-7}}2$, and is ***solvable in radicals*** whenever $\tau = \sqrt{-d}.$ Its seven roots $z_k$ in terms of the theta quotients $a,b,c$ are,

$$z_k = R_1(k)+h\, R_2(k)$$
$$R_1(k) = \zeta^{k}\,a^2+\zeta^{4k}\,b^2+\zeta^{2k}\,c^2$$
$$R_2(k) = \zeta^{6k}\,ab+\zeta^{3k}\,bc+\zeta^{5k}\,ca$$

where $\zeta = e^{2\pi i/7}$ and for $k=1,2\dots7.$ 


  [1]: https://mathworld.wolfram.com/j-Function.html