Relation between odd order theorem and the fact that every polynomial of odd order has at least one root

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It is known that there is connection between solvability of a group and expressing the roots of a polynomial by radicals, which is something I will study in this semester; however, by the odd order theorem

Every finite group of odd order is solvable.

However, I was wondering is there any relation between the fact that every polynomial of odd order has at least one root, with the fact that every finite group of odd order is solvable ?

Edit:

Since I haven't studied the main subject in detail (we just started in Ring theory in the graduate Algebra course that I'm taking)