As others have said, this can be answered with Galois theory (though the original proof predates Galois theory). Wikipedia's outline doesn't really describe a crucial step (equivalence of solvable groups and solvable by radicals) , so I'll try to give a few details here (I can't guarantee I'll succeed).
For an equation to be solvable with radicals it means we can start with the field of rational numbers, Q, and then construct larger fields by taking roots of elements in Q. For example, Q(√2) is the normal rationals extended to include solutions to x2 - 2. Every number in Q(√2) can be written as a + b√2, where a and b are rationals numbers. To solve, say, a 4th degree polynomial we may need to extend Q several times, e.g., Q ⊆ K1 ⊆ K2 ⊆ ... ⊆ Kn, but the point is that we can always do this in a way such that each field extension is the result of taking roots from elements in the previous field. To recap, a polynomial is solvable with radicals if it is a radical extension of Q, i.e., we can "reach" its solutions by iteratively taking roots.
Now, we need a little group theory. Every group G has a series G0 ⊆ G1 ⊆ ... ⊆ G associated with it, and we call G solvable if each quotient group Gi / Gi-1 is abelian (abelian means a+b = b+a). The series is sort of like the "factorization" of G by smaller groups.
This is where Galois theory comes into play. The fundamental theorem of Galois theory says that we can associate a sequence of groups G1 ⊆ G2 ⊆ ... ⊆ G with our (galois) sequence of fields K1 ⊆ K2 ⊆ ... ⊆ Kn. It turns out, if Ki / Ki-1 is a radical extension then the corresponding quotient group Gi / Gi-1 is an abelian group. Therefore, if every field extension is a radical extension then every corresponding quotient group is abelian. In particular, this means that having every field be a radical extension is therefore equivalent to G being a solvable group.
Finally, we can address your question. The question is, given a 5th degree polynomial f(x) = ax5 + bx4 + cx3 + dx2 + ex + g, do all the roots necessarily lie in a radical extension of Q? Well, with all polynomials we can associate a field extension and then compute its Galois group. The Galois group for this general f(x) is the symmetric group S5, and this group is not solvable because it has A5 / {1} in its series, and this group is not abelian.
To summarize this rough outline: For every polynomial, we can associate a series of field extensions. Galois theory tells us that this is a radical field extension (therefore solvable with radicals) if the Galois group is solvable. In the particular case of the general 5th degree polynomial f(x) given above, its Galois group is S5. This group is not solvable, so no general formula for solutions of 5th degree polynomials in radicals can exist.
I realize this may not make a whole lot of sense if you aren't already familiar with groups or fields, but I wanted to fill in some details that I think are missing or unclear from wikipedia.