It took almost 20 long years (I first found the quintic back in April 2006), but I finally have the answer. Given the two similar one-parameter forms,
$$u^5-10\alpha u^3+45\alpha^2u-\alpha^2 = 0\tag1$$
$$v^5-5\beta v^3+10\beta^2v-\beta^2 = 0\tag2$$
The first is well-known for its connection to the icosahedron, Rogers-Ramanujan continued fraction $R(q)$, and the McKay-Thompson series of class $1A$ for the Monster.
The second (it turns out) is more for the tetrahedron, Ramanujan's cubic continued fraction $C(q)$, and the McKay-Thompson series of class $6A$ for the Monster.
I. Definitions
Let $q=e^{2\pi i \tau}$ and Dedekind eta function $\eta(\tau)$, then,
$$C(\tau) = \cfrac{q^{1/3}}{1 + \cfrac{q+q^2}{1 + \cfrac{q^2+q^4}{1 + \ddots}}}$$
Define the eta quotients,
$$B(\tau) = \frac{\eta(\tau)\,\eta^2(6\tau)}{\eta(3\tau)\,\eta^2(2\tau)},\quad C(\tau)\, = \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}$$
as well as,
$$D_1(\tau) = \frac{\eta(\tau)\,\eta(3\tau)}{\eta(2\tau)\,\eta(6\tau)},\quad
D_2(\tau) = \frac{\eta(\tau)\,\eta(2\tau)}{\eta(3\tau)\,\eta(6\tau)} $$
Suppressing $\tau$ for brevity, we have the neat relations,
\begin{align}
\frac1{B^4}-\frac1{C^3} &= 1\\[6pt]
\frac1{B^4}+9B^4 &= D_1^6+10\\[6pt]
\frac1{C^3}-8C^3 &= D_2^4+7
\end{align}
Let $D_3=\frac{\eta(\tau)}{\eta(5\tau)}$ which is the McKay-Thompson series of class 5B for the Monster. These are then analogous to Ramanujan's well-known,
$$\frac1{R^5}-R^5 \,=\, D_3^6+11$$
but are McKay-Thompson series of class 6. The quotient $D_1^6$ is class 6C, $D_2^4$ is class 6D, while $\dfrac1{B^4}$ and $\dfrac1{C^3}$ are variants of class 6E (being A128633 and A105559). The forms $C$ and $D_2$ can solve the Bring quintic in a previous MO post, but $B$ and $D_1$ will be used in this one.
II. Modular equation
The modular equation between $B(\tau)$ and $B(5\tau)$ is more aesthetic,
$$\left(\frac{u^3}{v^3}-\frac{v^3}{u^3}\right)-5\left(\frac{u^2}{v^2}+\frac{v^2}{u^2}\right)+5\left(\frac{u}{v}-\frac{v}{u}\right)+3\left(\frac{1}{3u^2v^2}+3u^2v^2\right)=0$$
If $v = B(\tau)$, then the six roots $u$ are,
$$u = B(5\tau),\;-B\big(\tau/5\big),\;B\big(\tfrac{\tau+2}{5}\big),\;-B\big(\tfrac{\tau+4}{5}\big),\;B\big(\tfrac{\tau+6}{5}\big),\;-B\big(\tfrac{\tau+8}{5}\big)$$
Define,
$$X(\tau) = \left(\frac{B(5\tau)+B(\tau/5)}{\sqrt5\, B^3(\tau)\,D_1^3(\tau)}\right) \left(B\big(\tfrac{\tau+2}{5}\big)+B\big(\tfrac{\tau+8}{5}\big)\right) \left(B\big(\tfrac{\tau+4}{5}\big)+B\big(\tfrac{\tau+6}{5}\big)\right)$$
This function can solve the quintic.
III. Connection to cubic continued fraction
With scaling and change of variable, the equation
$$v^5-5\beta v^3+10\beta^2v-\beta^2 = 0$$
can be transformed to the simpler,
$$x^5+5x^3+10x+d=0$$
On the other hand, using the function $X(\tau)$ above, expand,
$$\prod_{n=0}^4 \big(x-X(\tau+2n)\big) = x^5+5x^3+10x+\sqrt{j_{6A}(\tau)-32\,}=0$$
where $j_{6A}(\tau)=\left(D_1^3(\tau)+\dfrac8{D_1^3(\tau)}\right)^2$ and is class 6A. Given $d$, one has to solve for $\tau$ in,
$$\sqrt{j_{6A}(\tau)-32\,} = d$$
which can be done. The roots $x_n$ for $n=0,1,2,3,4,$ are then,
$$x_n = X(\tau+2n)$$
$$\tau = \frac{_2F_1\big(\tfrac13,\tfrac23,1,\,1-\mu\big)}{_2F_1\big(\tfrac13,\tfrac23,1,\,\mu\big)}\sqrt{-\frac13}$$
$$\mu= \frac12\left(1-\sqrt{1-\frac{108}{j_3}}\right)$$
$$j_3 = \frac{432+45d^2+d^4-d(27+d^2)\sqrt{32+d^2}}2$$
Note: For simplicity, we used Ramanujan's theory of elliptic functions for signature $3$.
IV. Connection to tetrahedron
Given Ramanujan's cubic continued fraction $C=C(\tau)$. Define $z=-\dfrac{4C^2+C^{-1}}{3\sqrt2}$. (One can also define $z$ in terms of $B$.) Then a formula for the j-function is,
$$j(\tau) = \frac{-12^3(z^4-2\sqrt2\,z)^3}{(2\sqrt2\,z^3+1)^3}$$
However, these are polynomial invariants of the tetrahedral equation,
\begin{align}
&a :=z^4-2\sqrt2\,z\\
&b :=2\sqrt2\,z^3+1\\
&c :=z^6+5\sqrt2\,z^3-1
\end{align}
such that,
$$a^3+b^3=c^2$$
V. Connection to McKay-Thompson series 6A
Recall that,
$$x^5+5x^3+10x+\sqrt{j_{6A}(\tau)-32\,}=0$$
with $j_{6A}(\tau)$ as defined above. Then for $\tau=\sqrt{-d}$, the quintic has a solvable Galois group. For example, $j_{6A}\big(\sqrt{-5/6}\big)=320$. Then,
$$x^5+5x^3+10x+\sqrt{288}=0$$
is solvable in radicals.
VI. Tschirnhausen transformations
A. Given the Brioschi,
$$u^5-10c u^3+45c^2u-c^2 = 0$$
and the transformation,
$$x = \frac{au+b}{c^{-1}u^2-\color{blue}3}\quad$$
with unknowns $(a,b,c)$ to eliminate $u$ using resultants transforms the Brioschi to the form $x^5+f_1x^2+f_2x+f_3=0$. Equating this to a principal quintic $x^5+5px^2+5qx+r=0$, then one has a system of 3 equations in 3 unknowns where the final equation in $b$ has a deg-$2$ factor.
B. Given the quintic,
$$v^5-5c v^3+10c^2v-c^2 = 0$$
and the transformation,
$$x = \frac{av+b}{c^{-1}v^2-\color{blue}2},\quad c = -\frac{a b+2b^2}{4a^2}$$
transforms said quintic into Euler form $x^5+f_1x^2+f_2=0$. Equating this to $x^5+5px^2+q=0$, one has a system of 2 equations in 2 unknowns $(a,b)$ where the final equation in $b$ has a deg-$3$ factor.
Or if we do the transformation,
$$x = \frac{av+b}{c^{-1}v^2-\color{blue}1},\quad c = -\frac{a b+3b^2}{3a^2}$$
transforms said quintic into Bring form $x^5+f_1x+f_2=0$. Equating this to $x^5+5px+q=0$, one again has a system of 2 equations in 2 unknowns $(a,b)$ where the final equation in $b$ has a deg-$4$ factor.
