Consider the polynomial q(x) = x5-10x+2 over the rationals. This is irreducible by eisenstein's criterion. It is also a separable polynomial, since the rationals have characteristic 0. We will show that it's splitting field F, and its galois group G, are not solvable.
Since any root can be moved to any other root, the galois group is transitive. The group, acting on the roots of q(x), has one orbit, and |G| is divisible by the size of this orbit. This means |G| is divisible by 5.
Apply macay's theorem, and G contains the subgroup Z5. So G acts on 5 elements, the roots of q(x), and G includes a 5 cycle that shifts these elements around.
Use a little calculus in the xy plane to show the polynomial attains a maximum at -2¼ and a minimum at 2¼. These are all the real roots of q′(x), and there are no other local maxima or minima.
The graph crosses the x axis once to the left of -2¼, once between -2¼ and 2¼, and once to the right of 2¼. Therefore q(x) has 3 real roots and two complex roots.
Complex conjugation is an automorphism that swaps the two complex roots, and leaves the three real roots alone. This is known as a transposition. when a group includes a p-cycle and a transposition that group is equal to Sp. Therefore our galois group is S5, which is not solvable. It is not possible to describe the roots of q(x) using traditional operators.
In contrast, x5-2 is indeed solvable. Adjoin the fifth roots of 1 first, giving a cyclotomic extension with galois group Z4. Then adjoin the fifth root of 2, giving a second galois extension with galois group Z5. You can shift the roots of 1 while leaving the (real) fifth root of 2 alone, thus the cyclotomic automorphisms extend to the entire ring. The composite extension has a galois group whose order is at least 20; and the extension has dimension 20. This is galois, and solvable. Hence the quintic is solvable. Not surprising really, since one of its roots is the fifth root of 2, and that leaves a quartic behind, which is solvable.