Quadratics have the quadratic formula. Cubics have Cardano's formula (1545), and quartics have Ferrari's. Each one writes the roots using the coefficients, $+, -, \times, \div$, and $n$-th roots. For the quintic, people searched for almost three hundred years. Galois explained why they never found one: a polynomial's roots can be written with radicals exactly when its Galois group can be broken down into Abelian pieces, and $S_5$ can't.
"Writing a root with radicals" means building a tower where each floor adjoins one $m$-th root of something on the floor below.
A polynomial $f(x) \in \mathbb{Q}[x]$ is solvable by radicals if there exists a sequence of fields $\mathbb{Q} = F_0 \subseteq F_1 \subseteq \dots \subseteq F_k$ such that:
Show that $x^2 + bx + c$ (with $b, c \in \mathbb{Q}$) and $x^3 - 2$ are solvable by radicals.
Quadratic: the roots are $\frac{-b \pm \sqrt{b^2 - 4c}}{2}$, so the tower $\mathbb{Q} \subseteq \mathbb{Q}(\alpha_1)$ with $\alpha_1 = \sqrt{b^2 - 4c}$, $\alpha_1^2 \in \mathbb{Q}$, contains the splitting field. That's the quadratic formula, rewritten as a tower.
$x^3 - 2$: take $\alpha_1 = \zeta_3$ ($\alpha_1^3 = 1 \in \mathbb{Q}$) and $\alpha_2 = \sqrt[3]2$ ($\alpha_2^3 = 2 \in \mathbb{Q}(\zeta_3)$). Then $F_2 = \mathbb{Q}(\zeta_3, \sqrt[3]2)$ is the splitting field.
Notice the first floor of the $x^3 - 2$ tower: a root of unity. Radical towers almost always need roots of unity adjoined early, because $\sqrt[m]{a}$ alone only gives you one of the $m$ roots of $x^m - a$; the others are $\zeta_m^j \sqrt[m]{a}$. The proof of Galois' theorem below starts the same way.
The group-theory side of the story. How far is a group from being Abelian? Measure it with commutators: $[a, b] = aba^{-1}b^{-1}$ is $1$ exactly when $a$ and $b$ commute.
The commutator subgroup $G'$ of a group $G$ is the subgroup generated by all commutators $[a, b] = aba^{-1}b^{-1}$, $a, b \in G$.
$G' = \{1\}$ exactly when $G$ is Abelian. $G'$ is always normal, and $G/G'$ is always Abelian (in the quotient, $\overline{a}\,\overline{b} = \overline{b}\,\overline{a}$ because $aba^{-1}b^{-1} \in G'$). In fact $G'$ is the smallest normal subgroup with an Abelian quotient. Now repeat.
The derived series of $G$ is $G^{(0)} = G$, $G^{(i+1)} = \big(G^{(i)}\big)'$: $\ G \supseteq G' \supseteq G'' \supseteq \cdots$.
A group $G$ is solvable if there exists a sequence of subgroups $\{\mathrm{id}\} = H_k \le H_{k-1} \le \dots \le H_0 = G$ such that $H_i \trianglelefteq H_{i-1}$ and the quotient $H_{i-1}/H_i$ is Abelian for all $i$. Equivalently, its derived series eventually reaches the trivial group.
(Why equivalent: if the derived series hits $\{1\}$, it is such a sequence. Conversely, given such a sequence, $G/H_1$ Abelian forces $G' \le H_1$, then $G'' \le H_1' \le H_2$, and so on, so $G^{(k)} \le H_k = \{1\}$.)
Run the derived series on some groups you know. Each step replaces the group by its commutator subgroup; solvable groups shrink to $\{1\}$, and the others get stuck.
$S_3' = A_3$ and $S_4' = A_4$: the commutator of two permutations is always even, which you can check with signs from Symmetric Groups.
Two closure facts that the proof below needs (standard, and we'll use them without proof): subgroups and quotients of solvable groups are solvable, and if $N \trianglelefteq G$ with $N$ and $G/N$ both solvable, then $G$ is solvable.
The two halves meet.
A polynomial $f(x) \in \mathbb{Q}[x]$ is solvable by radicals if and only if the Galois group $\operatorname{Gal}(K/\mathbb{Q})$ of its splitting field $K$ is a solvable group.
The proof of "radicals $\implies$ solvable" translates the tower of fields into a chain of subgroups with the Fundamental Theorem. It uses one more lemma (part of Kummer theory, not proved here): if a field $L$ contains the $m$-th roots of unity and $a \in L$, then $L(\sqrt[m]a)/L$ is Galois with an Abelian (in fact cyclic) Galois group.
Proof outline (radicals $\implies$ solvable). Let $K$ be the splitting field of a polynomial $f \in \mathbb{Q}[x]$ which is solvable by radicals. Then there exists a tower $\mathbb{Q} = F_0 \subseteq F_1 \subseteq \dots \subseteq F_k$ such that $K \subseteq F_k$, where $F_i = F_{i-1}(\alpha_i)$ and $\alpha_i^{m_i} \in F_{i-1}$.
Step 1: adjoin roots of unity first. Let $m = \operatorname{lcm}(m_1, \dots, m_k)$, and define a new tower: $L_0 = \mathbb{Q}(\zeta_m)$, and $L_i = L_{i-1}(\alpha_i)$. Then $L_k \supseteq F_k \supseteq K$. We may also assume $L_k/\mathbb{Q}$ is Galois: if not, adjoin every conjugate of every $\alpha_i$ as well (each conjugate's $m_i$-th power lies in the corresponding lower floor of a conjugate tower), which keeps every step "adjoin an $m$-th root" and makes the top field a splitting field.
Step 2: every floor is Abelian. $L_0/\mathbb{Q}$ is Galois with Abelian Galois group $\cong (\mathbb{Z}/m\mathbb{Z})^\times$ (the Spicy Bonus on Galois Theory). Each $L_{i-1}$ contains $\zeta_m$, hence the $m_i$-th roots of unity, so by the lemma each step $L_i/L_{i-1}$ is Galois with an Abelian Galois group.
Step 3: translate to groups. Let $G = \operatorname{Gal}(L_k/\mathbb{Q})$ and $H_i = \operatorname{Gal}(L_k/L_i)$. The subgroups form a chain $\{\mathrm{id}\} = H_k \le H_{k-1} \le \dots \le H_0 \le G$. Applying the Fundamental Theorem to the Galois extension $L_k/L_{i-1}$, whose intermediate field $L_i$ is Galois over $L_{i-1}$: $H_i \trianglelefteq H_{i-1}$ and $H_{i-1}/H_i \cong \operatorname{Gal}(L_i/L_{i-1})$, which is Abelian. Likewise $H_0 \trianglelefteq G$ with $G/H_0 \cong \operatorname{Gal}(L_0/\mathbb{Q})$ Abelian. So $G$ is solvable.
Step 4: pass down to $K$. $K/\mathbb{Q}$ is Galois (a splitting field) and $K \subseteq L_k$, so by part 3 of the Fundamental Theorem, $\operatorname{Gal}(K/\mathbb{Q}) \cong G/\operatorname{Gal}(L_k/K)$ is a quotient of the solvable group $G$, and therefore is itself solvable. $\blacksquare$
(The converse, solvable $\implies$ radicals, also holds; it runs the same correspondence backwards, again with Kummer theory.)
A note on the notes: their outline writes $H_{i-1}/H_i \cong \operatorname{Gal}(L_{i-1}/L_{i-2})$; the indices should be $\operatorname{Gal}(L_i/L_{i-1})$. It also uses "$G = \operatorname{Gal}(L_k/\mathbb{Q})$" without checking that $L_k/\mathbb{Q}$ is Galois, which Step 1 above patches by adjoining conjugates.
Everything is in place for the punchline.
Show that $f(x) = x^5 - 4x + 2$ is not solvable by radicals.
The splitting field of $f$ has Galois group $G \cong S_5$ (Example 4 on Galois Theory: irreducible, three real roots).
The commutator subgroup of $S_5$ is $S_5' = A_5$: commutators are even, so $S_5' \le A_5$; and $A_5$ is a non-Abelian simple group, so its commutator subgroup (a normal subgroup of $A_5$, not $\{1\}$ because $A_5$ isn't Abelian) is itself: $A_5' = A_5$. Since $A_5 = A_5' \le S_5'$, we get $S_5' = A_5$.
So the derived series is $S_5 \supseteq A_5 \supseteq A_5 \supseteq A_5 \supseteq \cdots$ and never reaches the trivial group, meaning $S_5$ is not solvable. Therefore $x^5 - 4x + 2$ is not solvable by radicals.
And that kills the dream of a quintic formula: a general formula for degree $5$ (or higher, after multiplying by a power of $x$) would in particular express the roots of $x^5 - 4x + 2$ in radicals.
"No general formula" is not "no quintic can be solved." $x^5 - 2$ is solvable by radicals: its roots are $\zeta_5^j\sqrt[5]2$, built in the tower $\mathbb{Q} \subseteq \mathbb{Q}(\zeta_5) \subseteq \mathbb{Q}(\zeta_5, \sqrt[5]2)$. Its Galois group has order $20$ and is solvable. The theorem only says some quintics (the ones with Galois group $S_5$ or $A_5$) can't be done.
The quadratic formula fits on a T-shirt. The cubic formula fits on a poster. The quartic formula fits on a very angry poster. The quintic formula fits nowhere, and now we know it's not because nobody tried hard enough.
Derived series and one more unsolvable quintic (in the Spicy Bonus).
Show that $D_4$ is solvable by computing its derived series. Conclude that $x^4 - 2$ is solvable by radicals, and give a radical tower for it.
With $D_4 = \langle r, s : r^4 = s^2 = 1, srs = r^{-1}\rangle$: $[r, s] = rsr^{-1}s^{-1} = r \cdot (sr^{-1}s) = r \cdot r = r^2$. Every commutator is $1$ or $r^2$ (the quotient by $\langle r^2\rangle$ has order $4$, so is Abelian), so $D_4' = \langle r^2 \rangle \cong C_2$, which is Abelian, so $D_4'' = \{1\}$. Solvable.
So $x^4 - 2$, with Galois group $D_4$, is solvable by radicals. A tower: $\mathbb{Q} \subseteq \mathbb{Q}(i) \subseteq \mathbb{Q}(i, \sqrt[4]2)$, with $i^2 = -1 \in \mathbb{Q}$ and $(\sqrt[4]2)^4 = 2$. The roots are $i^j\sqrt[4]2$.
Show that every Abelian group is solvable, and that $S_3$ and $S_4$ are solvable. (You may use $S_4' = A_4$ and $A_4' = V_4$, as in the widget.)
Abelian: every commutator is $1$, so $G' = \{1\}$ right away.
$S_3$: $S_3' \le A_3$ (commutators are even), and $S_3$ isn't Abelian, so $S_3' \ne \{1\}$; $A_3$ has prime order, so $S_3' = A_3 \cong C_3$, Abelian, and $S_3'' = \{1\}$.
$S_4$: $S_4 \supseteq A_4 \supseteq V_4 \supseteq \{1\}$, and $V_4$ is Abelian, so $S_4''' = \{1\}$.
This is why the cubic and quartic formulas exist: every polynomial of degree $\le 4$ has a Galois group inside $S_4$, and subgroups of solvable groups are solvable.
Show that $x^5 - 2$ has a solvable Galois group by finding a chain of subgroups with Abelian quotients. (Hint: its splitting field is $\mathbb{Q}(\zeta_5, \sqrt[5]2)$, of degree $20$.)
Let $K = \mathbb{Q}(\zeta_5, \sqrt[5]2)$ and $G = \operatorname{Gal}(K/\mathbb{Q})$, $|G| = 20$. Let $N = \operatorname{Gal}(K/\mathbb{Q}(\zeta_5))$. Since $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ is Galois, $N \trianglelefteq G$ and $G/N \cong \operatorname{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q}) \cong C_4$: Abelian.
$|N| = [K:\mathbb{Q}(\zeta_5)] = 20/4 = 5$, so $N \cong C_5$: Abelian. The chain $\{1\} \trianglelefteq N \trianglelefteq G$ has Abelian quotients, so $G$ is solvable.
Build your own unsolvable quintic.
Prove that $x^5 - 6x + 3$ is not solvable by radicals.
Proof. Let $f(x) = x^5 - 6x + 3$, $K$ its splitting field, $G = \operatorname{Gal}(K/\mathbb{Q}) \le S_5$.
Irreducible: Eisenstein at $3$ ($3 \mid -6, 3$; $3 \nmid 1$; $9 \nmid 3$). So $5 \mid |G|$, and by Cauchy $G$ contains a $5$-cycle.
Exactly three real roots: $f(-2) = -17$, $f(-1) = 8$, $f(1) = -2$, $f(2) = 23$: sign changes on $(-2, -1)$, $(-1, 1)$, $(1, 2)$, so at least three real roots. $f'(x) = 5x^4 - 6$ has exactly two real zeros, so $f$ has at most two turning points and at most three real roots. The other two roots are a non-real conjugate pair, and complex conjugation acts on them as a transposition.
A subgroup of $S_5$ containing a $5$-cycle and a transposition is all of $S_5$, so $G \cong S_5$, which is not solvable. By Galois' theorem, $f$ is not solvable by radicals. $\blacksquare$