I know that not all quintics are solvable. But how do I identify the class of solvable ones?
Polynomials – How to Solve a Quintic Polynomial Equation
polynomialsroots
Related Solutions
For full details of this and more, the best place to look is the following paper:
D. S. Dummit, Solving solvable quintics. Math. Comp. 57 (1991), 387-401.
The main idea (which extends to any equation with a cyclic Galois group) is to consider Lagrange resolvents. Let the equation have roots $x_1,\ldots,x_5$ with an element $\tau$ of the Galois group permuting them as $x_i\mapsto x_{i+1}$. Let $\zeta=\exp(2\pi i/5)$ be the standard fifth root of unity. Then the Lagrange resolvents are
$\begin{align*} A_0&=x_1+x_2+x_3+x_4+x_5\\ A_1&=x_1+\zeta x_2+\zeta^2 x_3+\zeta^3 x_4+\zeta^4 x_5\\ A_2&=x_1+\zeta^2 x_2+\zeta^4 x_3+\zeta x_4+\zeta^3 x_5\\ A_3&=x_1+\zeta^3 x_2+\zeta x_3+\zeta^4 x_4+\zeta^2 x_5\\ A_4&=x_1+\zeta^4 x_2+\zeta^3 x_3+\zeta^2 x_4+\zeta x_5 \end{align*}$
Once one has $A_0,\ldots,A_4$ one easily gets $x_1,\ldots,x_5$. It's easy to find $A_0$ :-) The point is that $\tau$ takes $A_j$ to $\zeta^{-j}A_j$ and so takes $A_j^5$ to $A_j^5$. Thus $A_j^5$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in $A_1^5$ are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have $A_1$ as a fifth root of a certain explicit complex number. Then one can express the other $A_j$ in terms of $A_1$. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.
Galois Theory assigns, to each polynomial, a mathematical structure called a group. A polynomial is solvable in radicals (that is, you can write down its roots in terms of its coefficients, the 4 arithmetical operations, and square roots, cube roots, etc.) if and only if the corresponding group is a "solvable" group. The definition of solvable group won't mean much to you if you haven't done a course in group theory; there should be a sequence of groups, starting with the trivial group and ending with the group corresponding to the polynomial, such that each group in the sequence is a "normal" subgroup of the next group, and the "quotient" of each group by the previous group is "commutative".
What's more, if the group corresponding to the polynomial is solvable, then this sequence of in-between groups, together with the quotient groups, can be used to construct the formula for the roots of the polynomial.
To expand this into something you could actually use to determine whether a polynomial is solvable in radicals, and to solve it if it is, takes a semester of advanced undergraduate level mathematics. Get yourself a good text on Galois Theory (assuming you have already done courses on Linear Algebra and an introductory Abstract Algebra course - if not, you'll have to study those first), read it, and enjoy.
Best Answer
If you haven't reached that part of Galois theory yet, there is still a "simple" elementary way to test if a quintic is solvable. First, reduce it to depressed form (without the $x^4$ term).
Theorem (by Watson, 1930s): "Given an irreducible quintic with rational coefficients,
$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f = 0\tag1$$
If the sextic,
$$3125p^6 - 625(3c^2 + e)p^4 + 25(15c^4 + 8c d^2 - 2c^2e + 3e^2 - 2d f)p^2 + p\sqrt{D} + (-25c^6 - 40c^3d^2 - 16d^4 + 35c^4e + 28c d^2e - 11c^2e^2 + e^3 - 2c^2d f - 2d e f + c f^2)=0$$
and discriminant $D$,
$$D=-3200c^3d^2e^2 - 2160d^4e^2 + 6400c^4e^3 + 5760c d^2e^3 - 2560c^2e^4 + 256e^5 + 5120c^3d^3f + 3456d^5f - 11520c^4d e f - 10080c d^3e f + 4480c^2d e^2f - 640d e ^3f + 3456c^5f^2 + 2640c^2d^2f^2 - 1440c^3e f^2 + 360d^2e f^2 + 160c e^2f^2 - 120c d f^3 + f^4$$
has a root $p$ such that $p^2$ is rational, then $(1)$ is a solvable quintic."
See "Commentary on an unpublished lecture by G. N. Watson on solving the quintic" by Bruce Berndt. This is easily implemented in Mathematica and is useful when dealing with parametric quintics.
If you are in a rush and just want to determine if a particular equation is solvable, you can find the order of its Galois group using this online Magma calculator. For example, to test the solvable but irreducible $x^5-5x+12=0$, copy and paste the command,
One then finds the order is $10$, hence that quintic is solvable. All groups with order $<60$ are, though there are solvable groups with order $>60$.
Note: Don't forget the asterisk (*) between the numerical coefficient and the variable, like this: 5*x.
Addendum on Method 1:
We can reduce Method 1 to a simple rational root test by "root squaring".
Render the sextic equation above as
$\alpha p^6 + \beta p^4 + \gamma p^2 + p\sqrt{D} + \delta = 0$
Separate the odd degree term from the even degree ones:
$\alpha p^6 + \beta p^4 + \gamma p^2 + \delta = - p\sqrt{D}$
And square. Only even powers of $p$ now appear so we have a polynomial equation in $p^2$::
$(\alpha p^6 + \beta p^4 + \gamma p^2 + \delta)^2=p^2D$
$\alpha^2(p^2)^6 + 2\alpha \beta(p^2)^5 + (2\alpha \gamma + \beta^2)(p^2)^4 + 2(\alpha \delta + \beta \gamma)(p^2)^3 + (2\beta \delta + \gamma^2)(p^2)^2 + (2 \gamma \delta - D)(p^2) + \delta^2 = 0$
You may then apply the standard rational root test to this equation. If it passes, the conditions for solvability of the quintic are satisfied.