Worked solution: Abel–Ruffini theorem and the Galois solvability criterion (1824)

Take $f(x)=x^5-4x+2$. By Eisenstein's criterion at the prime $p=2$ (the leading coefficient $1$ is not divisible by $2$, every other coefficient $0,0,0,-4,2$ is divisible by $2$, and the constant term $2$ is not divisible by $2^2=4$), $f$ is irreducible over $\mathbb{Q}$, so its Galois group $G$ acts transitively on the $5$ roots and hence has order divisible by $5$; by Cauchy's theorem $G$ contains a $5$-cycle. Calculus pins down the real roots: $f'(x)=5x^4-4$ vanishes at only two real points, so by Rolle's theorem $f$ has at most $3$ real roots, and evaluating $f(-2)<0<f(0)$, $f(0)>0>f(1)$, $f(1)<0<f(2)$ shows (via the intermediate value theorem) it has at least $3$; so $f$ has exactly $3$ real roots and one complex-conjugate pair. Complex conjugation is then a field automorphism swapping just that pair, i.e. a transposition in $G$. A theorem of elementary group theory says a subgroup of $S_p$ ($p$ prime) containing a $p$-cycle and a transposition must be all of $S_p$, so $\mathrm{Gal}(f/\mathbb{Q})\cong S_5$ — the very group shown unsolvable in Step 7 — and by Galois's criterion (Step 5), $x^5-4x+2=0$ cannot be solved by any radical formula.