We may as well assume the polynomial is irreducible, for if not, we can always derive the roots of the irreducible factors in turn.
As we shall see, groups and field extensions are called "solvable" because they correspond to solvable polynomials. In fact, we will show that A separable polynomial q(x) is solvable iff its galois group is solvable.
First, a few comments on the conditions of the theorem.
I call the polynomial q(x) to avoid confusion with phrases like characteristic p, and pth root.
We assume q(x) is separable, so that its splitting field F is a galois extension. This takes care of polynomials over finite fields and fields with characteristic 0. Those are the fields you run in to most of the time anyways.
Any radical of composite order can be split into successive radicals of prime order, so assume only pth roots are extracted for various primes p. If you want the sixth root, take the cube root, then the square root.
If p is the characteristic of K, extracting a pth root in the literal sense either produces another element in K and can be ignored, or it produces a purely inseparable extension of the previous intermediate field. Thus the pth root lies in an inseparable subfield of F, F is not separable, and q(x) is not separable. Therefore we assume a pth root of a over K, with characteristic p, implies a procedure to solve the polynomial xp-x-a. That's what we mean by a pth root when K has characteristic p. This abuses the notation a bit, but it makes the theorem work.
Now assume all the roots of q(x) are expressible using field operations, radicals, and elements of the base field K. build a finite extension of K that contains the first root, then adjoin the second, then the third, and so on. Finally you have a field extension that contains all the roots of q(x). With this in mind, let's take the roots one at a time.
Select any root of q(x). Remember, this root is expressed as a sequence of field operations and radicals.
If two elements have been adjoined, i.e. brought into a field extension of K, these can be added, multiplied, etc, and the result lies in the same field. The field only increases when we apply the radical operator.
Let z be a member of the field F, which could already be an extension of K, and look for a field extension E/F that contains the pth root of z. If this root is already in F we are done. Otherwise consider two cases.
If p agrees with the characteristic of F, which is the same as the characteristic of K, simply adjoin the pth root of z. Remember that this "root" is actually a root of the polynomial xp-x-z = 0. When this root is adjoined, the extension E/F is cyclic by an earlier theorem.
Next assume p is not equal to the characteristic of K. Adjoin the pth roots of 1, which is a cyclic extension. Then adjoin the pth root of z. This too is a cyclic extension by the previous theorem.
Build a tower of cyclic extensions as operators and radicals are applied. The result is a solvable extension that contains the specified root of q(x). Do this for all the roots in q, and the splitting field for q is contained in a solvable extension. Since q is separable its splitting field is galois. The splitting field of q is a normal field extension inside the solvable extension we have created. Its galois group is a quotient of the solvable galois group. The quotient of a solvable group is solvable, hence the galois group associated with the polynomial q(x) is solvable.
Now for the converse. Let the splitting field E have a solvable galois group G. Build a chain of normal subgroups, and a tower of normal extensions, that takes you from K all the way up to E. If every element in E can be derived using field operators and radicals, then the roots can be so derived, and q is indeed solvable.
Let the chain up subgroups of G, and the tower of intermediate extensions of E, have height n. Assume, by induction, every element in a shorter tower is expressible with field operators and radicals.
Let H be the last nontrivial subgroup in the solvable series of G. We have selected a maximal solvable series, so H = Zp for some prime p.
By definition, H is normal in the previous subgroup of G (which contains it). This does not mean H is normal in G. However, a beautiful theorem tells us that if G is a solvable series, then indeed each subgroup is normal in G. Therefore H is normal in G.
Let F be the intermediate field extension that is fixed by H. Since H is normal in G, F is a normal extension and a galois extension. It's galois group is G/H, which is solvable, and has a height of n-1 or less. Everything in F is accessible using our operators.
The extension E/F is galois, with galois group Zp, hence it is a cyclic extension.
We are interested in the element r, which lies in E, but not in F. Adjoining r cannot produce an intermediate extension, so F(r) = E. Adjoininr r produces a cyclic extension of order p.
Refer to the previous two theorems, when K does or does not have characteristic p. In either case, r can be expressed as the pth root (or radical) of z, for some z in F. Since z is also expressible using field operators and radicals, r can be derived. this holds for every r in E, including the roots of q(x).
Let's put this all together. Let q(x) be a separable polynomial, e.g. any polynomial over the reals. Now q can be "solved", its roots expressed using field operators and radicals, iff its splitting field is a solvable extension, iff its splitting field has a solvable galois group.