What a radical tower really is
A formula ‘in radicals’ uses only +, −, ×, ÷ and nth roots. Algebraically that means: starting from K you reach the solution inside a tower K = F_0 ⊆ F_1 ⊆ … ⊆ F_m where each step adjoins a single radical, F_{i+1} = F_i(α) with α^{n_i} ∈ F_i. An equation is solvable by radicals when its splitting field sits inside such a tower. The plan is to read this tower through the Galois correspondence and see what it forces on the group.
There is one technical sweetener. Adjoining α with α^n ∈ F_i is a cyclic extension *provided F_i already contains the nth roots of unity*. So first throw all needed roots of unity into the base — a cyclotomic step, itself abelian hence solvable — and afterward each radical step has cyclic Galois group. We can arrange the tower to be normal over K with each layer cyclic.
Tower of fields ↔ chain of subgroups
Now apply the correspondence to the normal tower K ⊆ F_1 ⊆ … ⊆ F_m. Each F_i corresponds to a subgroup G_i = Gal(F_m/F_i), and the inclusions flip to a chain G = G_0 ⊇ G_1 ⊇ … ⊇ G_m = {1}. Because each F_{i+1}/F_i is normal with cyclic group, each G_{i+1} ◁ G_i with cyclic quotient G_i/G_{i+1}. A group with such a chain — successive quotients abelian — is by definition a solvable group. That is the theorem: solvable by radicals ⇒ solvable Galois group.
A quintic that breaks
To kill solvability we need a quintic whose group is not solvable. The full symmetric group S_5 works: it has the chain S_5 ⊃ A_5 ⊃ {1}, but A_5 is a nonabelian simple group — no further proper normal subgroup, no abelian quotient available. So S_5 is not solvable, and any quintic with Galois group S_5 cannot be solved by radicals. We just need one concrete polynomial that lands there.
Claim: f(x) = x^5 - 6x + 3 over Q has Galois group S_5. Step 1 (irreducible). Eisenstein at p = 3: coeffs of x^4..x^0 are 0, 0, 0, -6, 3; all of -6, 3 divisible by 3, leading coeff 1 not, and 3 not divisible by 3^2 = 9. => f is irreducible over Q, so 5 | |Gal(f)| and the group is transitive. Step 2 (a 5-cycle). Transitive subgroup of S_5 of order divisible by 5 contains an element of order 5 (Cauchy) = a 5-cycle. Step 3 (a transposition). Count real roots via calculus: f'(x) = 5x^4 - 6 has two real critical points, so f has exactly 3 real roots and 1 complex-conjugate pair. Complex conjugation fixes the 3 real roots, swaps the 2 complex roots => it acts as a transposition in Gal(f) <= S_5. Step 4 (generation). A 5-cycle and ANY transposition together generate all of S_5. => Gal(f/Q) = S_5, which is NOT solvable. => x^5 - 6x + 3 = 0 cannot be solved by radicals.