I was studying for a Galois theory exam when I saw this question and was stumped. It is not hard to check that $f$ is irreducible over $\Bbb Q$ (since $x^5\cdot f(\frac{1}{x})$ is irreducible over $\Bbb Q$ [Eisenstein criterion $p=5$]).
Here is my main problem, most questions like this have a polynomial for which we can compute the roots exactly. But, in our case, $Gal(L/{\Bbb Q}) \cong S_5$ ($L\subseteq \Bbb C$ is the splitting field of $f$ over $\Bbb Q$) so we can't even express the roots in a useful way.
I tried taking a similar approach by trying to show $x^5\cdot f(\frac{1}{x})$ is irreducible over $\Bbb Q[\sqrt{3}]$, again by applying the Eisenstein criterion but now over $\Bbb Z[\sqrt{3}]$. However this would require knowing how $20,35,10$ factor in $\Bbb Z[\sqrt{3}]$ which is probably not the way to go.
Any help would be appreciated.
EDIT: I think this is a solution but I am not sure:
If $f$ splits over $\Bbb Q[\sqrt{3}]$ then one both of it's factors are of degree less or equal to 4, hence they are solvable by radicals but since their coefficients are in $\Bbb Q[\sqrt{3}]$ then the roots are expressable as radicals and so $f$ is solvable in radicals which is a contradiction to $Gal(L/{\Bbb Q}) \cong S_5$.
Is my solution correct?