solvability by radicals

A polynomial equation is solvable by radicals if you can write its roots using only the everyday ingredients — the numbers you started with, the four arithmetic operations, and root extractions (square roots, cube roots, and so on) nested as deeply as you like. The quadratic formula is the famous example: its roots come from one square root applied to the coefficients.

Precisely, an equation over a field K is solvable by radicals if its roots lie in some tower of extensions K = K0 ⊆ K1 ⊆ ... ⊆ Kn where each step Ki+1 is obtained from Ki by adjoining an n-th root of an existing element. Galois's theorem makes this condition group-theoretic: the equation is solvable by radicals if and only if its Galois group is a solvable group — a group that can be broken down into abelian (commutative) pieces.

This is a statement about formulas, not about whether roots exist. By the fundamental theorem of algebra every polynomial of degree at least one has roots in the complex numbers. Solvability by radicals asks the sharper question: can those roots be expressed by a finite formula built only from arithmetic and radicals? For the general quintic and beyond, the answer is no.