Finding Galois group of a separable polynomial The question asks the following:

Let $E$ be the splitting field of $x^4-10x^2-20$ over $\mathbb{Q}$. Find $Gal(E/\mathbb{Q})$. 

Since this polynomial is irreducible by Eisenstein's criterion, we know that the Galois group acts transitively on the set of four distinct roots of this polynomial. However, after that, I am very lost as to how should I proceed. One approach which I thought of is to check the transitive subgroups of $S_4$, but that seems to make things more complicated. I would appreciate hints on this problem. 
 A: The resolvent cubic of the quartic is $$C(x) = x^3+20x^2+180x=x(x^2+20x+180)$$Note that the discriminant of the quadratic is $20^2-4\cdot180=-320$, so $C(x)$ factors into a linear and quadratic term.
The discriminant of the cubic is $D=20^2180^2-4\cdot180^3=160^2\cdot5$
The original quartic is reducible in $\mathbb Q(\sqrt D)=\mathbb Q(\sqrt5)$ since it can be written as $(x^2-5)^2-45=(x^2-5+3\sqrt5)(x^2-5-3\sqrt5)$
So, the Galois group is $D_8$.
A: Let $\alpha_1=\sqrt{5+3\sqrt{5}},\alpha_2=\sqrt{5-3\sqrt{5}}$, then the splitting field of $f$ is $E=\mathbf{Q}(\alpha_1,\alpha_2)$. As $f$ is irreducible, we know that $\alpha_1$ has degree $4$ over $\mathbf{Q}$. In particular, $\mathbf{Q}(\alpha_1)$ and $\mathbf{Q}(\alpha_2)$ are quadratic extensions of $\mathbf{Q}(\sqrt{5})$. They are equal if and only if $\alpha_1\alpha_2=\sqrt{-20}\in \mathbf{Q}(\sqrt{5})$, which is not the case (see comment in this answer). Now, $\alpha_2$ is a zero of $X^2+\alpha_1^2-10\in \mathbf{Q}(\alpha_1)[X]$, so $$|\operatorname{Gal}(E/\mathbf Q)|=[E:\mathbf{Q}]=[\Omega:\mathbf{Q}(\alpha_1)][\mathbf{Q}(\alpha_1):\mathbf{Q}]=2\cdot 4=8.$$
As $D_8$ is the only transitive subgroup of order 8 of $S_4$, we conclude $\operatorname{Gal}(E/\mathbf{Q})\cong D_8$.
