Is there something like Cardano's method for a SOLVABLE quintic. So there is no quadratic formula equivalent for a GENERAL fifth degree equation, but is there an equivalent formula for a SOLVABLE fifth degree equation.

For any solvable polynomial equation $a_px^p+a_{p-1}x^{p-1}+\cdots+a_1x+a_0=0$ of odd prime degree $p$, the roots have the form:
$x=-\frac{a_{p-1}}{pa_p}+\omega\sqrt[p]{r_1}+\omega^2\sqrt[p]{r_2}+\cdots+\omega^{p-1}\sqrt[p]{r_{p-1}}$
Where $r_1,\cdots,r_{p-1}$ are solutions to a polynomial of degree $p-1$ with coefficients in the same base field as the original equation.
A to determine which polynomial, one method is to set each of these expressions equal to a different root and deriving elementary symmetric polynomials in the $r_i$, and solving a linear system.
Example: $x^5-5x+12$ with one real root and Galois Group $D_5$.
We know that the roots are
$x_1=a+b+c+d$
$x_2=a\text{ cis }\frac{2\pi}{5}+b\text{ cis }\frac{4\pi}{5}+c\text{ cis }\frac{6\pi}{5}+d\text{ cis }\frac{8\pi}{5}$
$x_3=a\text{ cis }\frac{4\pi}{5}+b\text{ cis }\frac{8\pi}{5}+c\text{ cis }\frac{2\pi}{5}+d\text{ cis }\frac{6\pi}{5}$
$x_4=a\text{ cis }\frac{6\pi}{5}+b\text{ cis }\frac{2\pi}{5}+c\text{ cis }\frac{8\pi}{5}+d\text{ cis }\frac{4\pi}{5}$
$x_5=a\text{ cis }\frac{8\pi}{5}+b\text{ cis }\frac{6\pi}{5}+c\text{ cis }\frac{4\pi}{5}+d\text{ cis }\frac{2\pi}{5}$
Here, $a,b,c,d$ are fifth-roots of solutions to a quartic.
If you ask Mathematica to solve the quintic in closed form, it will give you root objects. The first one is real, the next two are one complex conjugate pair, and the last two are the other complex conjugate pair.
The expressions of the pair $(x_1,x_4)$ are complex conjugates, and so are expressions for $(x_2,x_3)$. This gives you a limited number of possible candidates for which $x$ corresponds to which root object that Mathematica gave.
For each candidate assignment, ask mathematica to simplify fully with "FullSimplify", the elementary symmetric polynomials in $a^5,b^5,c^5,d^5$ and look for the case where only rational numbers are spat out. These are $\pm$ coefficients of the auxiliary quartic with $a^5,b^5,c^5,d^5$ as the roots. Mathematica may take a while in this step, so be patient.
In our case, the correct assignments are $x_1=r_1$, $x_2=r_3$, $x_3=r_4$, $x_4=r_5$, $x_5=r_2$.
Here, $r_i$ are the root objects Mathematica gave.
The elementary symmetric polynomials in $a^5,b^5,c^5,d^5$ come out:
$e_1=-4, e_2=\frac{4}{5}$, $e_3=\frac{8}{125}$, $e_4=-\frac{1}{3125}$.
The quartic: $u^4+4u^3+\frac{4}{5}u^2-\frac{8}{125}u-\frac{1}{3125}=0$
Solving it and matching up the expressions correctly to $a^5,b^5,c^5,d^5$, we are done.
Note that mathematica evaluates $n$th roots of negative real numbers to complex numbers with argument $\frac{2\pi}{n}$, instead of evaluating them to be equal to negative real when $n$ is odd. You must account for that to check your work by flipping sign on the entire expression in the fifth root and the sign in front of the fifth root if the inside expression is negative real.
Final answer (accounting Mathematica's $n$th root preferences):
$x= -\omega\sqrt[5]{1+\frac{2}{5}\sqrt{5}-\frac{3}{25}\sqrt{125+55\sqrt{5}}}+\omega^2\sqrt[5]{-1+\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125-55\sqrt{5}}}-\omega^3\sqrt[5]{1-\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125-55\sqrt{5}}}-\omega^4\sqrt[5]{1+\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125+55\sqrt{5}}}$
Theoretically, these steps can be repeated for any quintic, and replicated on any polynomial of odd prime degree. The only obstacle is in solving the auxiliary polynomial of degree $p-1$, beginning with $p=7$ onwards, which would require finding an intermediate field in which it factors. This can be done by assuming factorization and solving a system in Mathematica.
