How to solve decics like $x^{10}+100x^2+160x+64=0$ having Galois group 10T33?

A big difference to Smart way to solve octics like $x^8+5992704x-304129728=0$ is that in the prior example the field $E$ defined by the polynomial $f$ (of degree 8) had a subfield $S$ (of degree 4) over which it is quadratic. Thus finding the subfield reduced everyting to the degree $\le 4$ formulas.

Here you have two difficulties. The first is that -- while there still is a subfield -- the degree over this subfield is 5. So you cannot use a generic formula, but would need to go into the generic Galois theory machinery with resolvent, roots of unity etc. for degree $\ge 5$.

The second difficulty is that the $E$ as extension of $S$ will have Galois group not cyclic, but of order $20$. You thus will have to do three steps, corresponding to the composition series of the Frobenius group of order $20$.

This can be done (even without need to go to hypergeometric series, as Mathematica seems to do), but the resulting expression in radicals will, in all likeness, be a nightmare, and if you are just looking for the result "out of curiosity" most likely not be worth the effort.

To solve equations of Fermat prime degree $p=2^{n}+1$ with a solvable Galois group, one generally needs a resolvent equation of degree $2^n$ (which, using only nested square roots, can factor into quadratics). Thus to solve the solvable quintic,

$$x^5+10i x+8i=0$$

one employs 4 roots of a quartic (factored into 2 quadratics) which in this case nicely involves the golden ratio $\phi$. Let,

$$\begin{aligned}\color{red}u &= \frac{\sqrt{2\,i}}{\;5}\sqrt{1+2\,i}\\ &= \frac{\sqrt{2\,i}}{\;5}\left(\sqrt{\frac{1+\sqrt5}2}+\sqrt{\frac{1-\sqrt5}2}\right)\\ &= 0.0971+0.4116i\end{aligned}$$

(Note: All numerical values are generated by Mathematica and truncated.)

I. New answer

The five roots $x_k$ are then

$$x_k = u\,T+\frac1T-\frac{a}{T^2}+\frac{b}{T^3}$$

where $$T_k = \zeta^k \Big(\frac{ab}{u}\Big)^{1/5}$$

$$a=\frac1u\big(i-\sqrt{u-1}\big) = -0.05367 + 0.5009 i$$

$$b=\frac1u\big(-c+\sqrt{c^2+1}\big) = 0.24289 + 0.00326 i$$

$$\quad c = \frac{1-5i}2+\frac1u \;=\; 1.04322 - 4.8011i$$

with $\zeta = e^{2\pi i /5}$ and $k = 0,1,2,3,4$.

The advantage of this version is there is only one $z^{1/5}$ root extraction and does not involve the $5$th root of several complex numbers (and consequent ambiguity) as in the old answer below.

II. Old answer

Let $\color{red}u$ be as defined above. Then the resolvent quadratics are,

$$z^2+2u(5u+2i)z+u^5 = 0\\z^2+2u(5u-2i)z-u^5 = 0$$

So the roots $z_i$ are,

$$z_1 = -0.0657 - 0.3821 i\\ z_2=0.0191 - 0.0291i\\ z_3=0.0028 + 0.0027i\\ z_4=3.2437 - 1.1914i$$

Taking their $5$th roots and affixing a correct $5$th root of unity $\zeta = e^{2\pi i/5}$,

$$z_1^{1/5}\zeta^2 = -0.3464 + 0.3758i \\z_2^{1/5}\zeta^2 = -0.2934 + 0.1511i \\z_3^{1/5}\zeta^2 = -0.4632 + 0.6855i \\z_4^{1/5}\zeta^4 = 0.3092 - 1.2435i$$

The sum of those $5$th roots,

$$x_1 = z_1^{1/5}\zeta^2+z_2^{1/5}\zeta^2+z_3^{1/5}\zeta^2+z_4^{1/5}\zeta^4 = -0.7938 - 0.03104i$$

is then a root of the quintic,

$$x^5+10i x+8i=0$$

P.S. The other roots can be found by affixing appropriate $\zeta^k$ to the $z_i^{1/5}$.