In this interesting post, it quotes a paper that to write down the complete solution (not in radicals) of the general quintic, one would need "a piece of paper as big as a large asteroid". Probably an exaggeration, but perhaps we can reduce the size.
The quartic transformation that result in the Bring-Jerrard form can apply to any equation of deg $n>4$, and to undo everything later is a mess. But there is a transformation unique to the quintic which exploits its connection to the icosahedron and require only square roots, hence is much tidier.
I. From general to principal quintic
One first transforms the general quintic $P(x) = 0$ to principal form $P(y) = 0$ using a quadratic Tschirnhausen transformation $x^2+mx+n = y$. Thus, to recover the roots $x$ from $y$ need only solving a quadratic.
II. Solution of the principal quintic
Now that we have the principal form,
$$P(y)=y^5+5ay^2+5by+c = 0$$
then, like the Bring quintic, we can solve it using the Dedekind eta function $\eta(\tau)$ and hypergeometric function $_2F_1$ but without solving any cubic or quartic at all. Instead, the Rogers-Ramanujan continued fraction $R(q)$ appears. It can be done in just $9$ steps.
First, solve the quadratic for any root $\lambda$,
$$(a^4 + a b c - b^3)\lambda^2 - (11a^3b - a c^2 + 2b^2 c)\lambda + (64a^2b^2 - 27a^3c - b c^2) = 0\tag1$$
then define $e,f$ as,
$$e = \frac{-8a \lambda^3 - 72b \lambda^2 - 72c \lambda + g a^2}{a \lambda^2 + b \lambda + c} \tag2$$
$$f = \frac{1}{1728 - g} \tag3$$
$$g = \frac{(a \lambda^2 - 3b \lambda - 3c)^3}{a^2(a c \lambda - b^2\lambda - b c)} \tag4$$
Let $q = e^{2\pi i\tau}$ and the principal quintic roots $y$ are,
$$y = \frac{ez+\lambda}{f^{-1}z^2-3}\tag5$$
$$z = \pm\sqrt{\frac{-f(u^2 + 4)(u^2 - 2u - 4)^2}{u^5 + 5 u^3 + 5u - 11}}\tag{6a}$$
Or alternatively,
$$\quad z = \pm\left(\frac{1}{R(q)}+R(q)\right)\sqrt{\frac{-f(u^2 - 2u - 4)^2}{v - 11}}\tag{6b}$$
$$u = \frac{1}{R(q)}-R(q) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1\tag{7a}$$
$$\quad v = \frac{1}{R^5(q)}-R^5(q) = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11\tag{7b}$$
$$\qquad\qquad \tau=\tau_k = k+ \frac{_2F_1\big(\tfrac16,\tfrac56,1,1-\beta\big)}{_2F_1\big(\tfrac16,\tfrac56,1,\beta\big)}\sqrt{-1}, \quad (k = 0,1,2,3,4) \tag8$$
$$\qquad\qquad 4\beta^2-4\beta+\frac{1728f}{1728f-1} = 0\quad(\text{any root}\;\beta)\\ \tag9$$
Steps $1-5$ transform the principal quintic into the Brioschi as described in this post and can be found in Duke's and Tóth's "The Splitting of Primes in Division Fields of Elliptic Curves" in page 10. (Note: There is a stumbling block in Step 4 when $a=0$, but one can get around that by redefining the first four steps.) The steps afterwards are mine and a more efficient version of this 2015 post.
Thus, by passing through the Brioschi quintic, it seems one may not need an asteroid-sized piece of paper after all.
III. Example
Given,
$$y^5+5y^2+10y+2=0$$
so $a,b,c = 1,2,2.\,$ Solving $(1)$, we choose $v = \tfrac{1}{3}\big({-17} + \sqrt{871}\big)$ and get,
$$e = \tfrac{16}{27}\big(5071 - 179\sqrt{871}\big)$$
$$f = \tfrac{1}{43312^2}\big(615193 + 20894 \sqrt{871}\big)$$
Substituting those into Steps $5$-$9$, and choosing the correct sign of $z$, we find the five roots $y$.
IV. Question
Note that Step $6\text{a}$ solves the Brioschi quintic,
$$z^5-10fz^3+45f^2z-f^2 = 0$$
$$z = \pm\sqrt{\frac{-f(u^2 + 4)(u^2 - 2u - 4)^2}{u^5 + 5 u^3 + 5u - 11}}\tag{6a}$$
Q. But is there a way to simplify the expression in $u$ under the square root sign? Or solve for $z$ without square roots at all? (The polynomials are so familiar I feel the solution must already have been tackled by Brioschi, Klein, or their contemporaries.)
Update: As pointed out by @dxiv and @ParamanandSingh in the comments, we can indeed simplify the root $z$ as,
$$\quad z = \pm\sqrt{-f\;}\,(u^2 - 2u - 4)\left(\frac{1}{R(q)}+R(q)\right) \left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{3}\tag{6b}$$
where $u = \frac{1}{R(q)}-R(q) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1,$ so only $f$ remains under the square root, confirming my belief that $z$ must have a simpler form. (But this can't be the solution found by Brioschi or his contemporaries since they lived before Ramanujan.)
