Show that the polynomial of $x^7-10x^5+15x+5$ is not solvable by radicals.
For a polynomial of degree 5, we simply consider the derivative and determine the number of real and complex roots from there. However, with a degree 7 polynomial, this becomes much more difficult.
Here's an outline of one possible approach, and I assume you're working over $\mathbb{Q}$. This polynomial, call it $f$, is $5$-Eisenstein, hence irreducible over $\mathbb{Q}$. Now use Descartes' rule of signs to show there are at most $5$ real roots. You can show there actually exist $5$ real roots by the intermediate value theorem by exhibiting $5$ disjoint intervals over which the polynomial changes sign, indeed $(3,4),(1,2),(-1,0),(-2,-1),(-4,-3)$, should work.
So $f$ is an irreducible degree $7$ polynomial with exactly two nonreal roots. Hence its Galois group is all of $S_7$.
It's known from group theory that $S_7$ is not solvable, so $f$ is not solvable by radicals.