Call $K= \mathbb{F}_5[x]/(x^2+3x+4)$. Then the polynomial $x^2+3x+4 \in \mathbb{F}_5[x]$ has a root $ \alpha \in K$. But now, one can write
$$x^2+3x+4 = (x- \alpha)g(x)$$
for some $g \in K[x]$. Since the degree of $x^2+3x+4$ is 2 and the degree of $x- \alpha$ is 1, $g$ must be a polynomial of degree 1, so it has a root $\beta \in K$.
Now, $x^2+3x+4$ can have at most two roots, so $\alpha, \beta \in K$ are all of them.
Factorizing the polynomial gives $(x^4 - 8x^2 - 5)(x + 1)(x - 1)$.
Hence the splitting field $L/\mathbb Q$ is the same as the splitting field of $g(x) = x^4 - 8x^2 - 5 = (x^2 - 4)^2 - 21$.
If $\alpha \in L$ is a zero of $g(x)$, then we have $(\alpha - 4)^2 = 21$, hence $L$ contains the subfield $K = \mathbb Q[\sqrt{21}]$.
It remains to factorize the polynomials $x^2 - (4 \pm \sqrt{21})$ over the field $K$. This is the same as checking whether $4 \pm \sqrt{21}$ is a square in $K$.
We do this by calculating the norm $N_{K/\mathbb Q}(4 \pm \sqrt{21}) = -5$, which is not a square in $\mathbb Q$, hence $4 \pm \sqrt{21}$ are not squares in $K$.
Therefore, $L$ contains the quadratic extension $E = K[\sqrt{4 + \sqrt{21}}]$ of $K$, which has degree $4$ over $\mathbb Q$.
It remains to factorize $x^2 - (4 - \sqrt{21})$ over $E$, i.e. to check whether $4 - \sqrt{21}$ is a square in $E$.
We prove a simple lemma.
Lemma: Let $F$ be a field of characteristic different from $2$, and $a, b$ be non-square elements of $F$. Then $b$ is a square in the extension $F[\sqrt a]$ if and only if $ab$ is a square in $F$.
Proof: The if part is clear. For the only if part, suppose $b = (u + v\sqrt a)^2$, we then have $b = u^2 + v^2a + 2uv\sqrt a$. Since $1, \sqrt a$ form an $F$-basis of $F[\sqrt a]$, it follows that $uv = 0$. But $v = 0$ implies that $b$ is a square in $F$, so we have $u = 0$ and $b = v^2a$, hence $ab = (va)^2$ is a square in $F$.
Now in our problem, applying the lemma to $F = K$ and $a, b = 4 \pm \sqrt{21}$, we see that it suffices to check whether the product $(4 + \sqrt{21})(4 - \sqrt{21}) = -5$ is a square in $K$.
Since $K = \mathbb Q[\sqrt{21}]$, we apply again the lemma to $F = \mathbb Q$ and $a, b = 21, -5$, and the problem becomes to check whether $21 \times (-5) = -105$ is a square in $\mathbb Q$. As it certainly is not, we conclude that $4 - \sqrt{21}$ is not a square in $E$.
Thus the splitting field $L$ is the quadratic extension $E[\sqrt{4 - \sqrt{21}}]$ of $E$, which has degree $8$ over $\mathbb Q$.
Putting together, one obtains $L = \mathbb Q[\sqrt{21}, \sqrt{4 + \sqrt{21}}, \sqrt{4 - \sqrt{21}}]$. It is easy to verify that it can also be written as $L = \mathbb Q[\sqrt{-5}, \sqrt{4 + \sqrt{21}}]$.
Best Answer
Let $a=\sqrt{2}$, and let $b=\sqrt{-3}$.
Let $\omega=\exp\left(i\left({\large{\frac{\pi}{3}}}\right)\right)$.
Noting that
it follows that the roots of $x^6-8$ in $\mathbb{C}$ are $$a,a\omega,a\omega^2,a\omega^3,a\omega^4,a\omega^5$$ hence $K=\mathbb{Q}(a,\omega)$ is a splitting field of $f$.
Then since $$ \omega = \cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right) = \frac{1}{2}+\frac{1}{2}i\sqrt{3} = \frac{1}{2}+\frac{1}{2}b $$ it follows that we can also write $K=\mathbb{Q}(a,b)$, or explicitly, $K=\mathbb{Q}\left(\sqrt{2},\sqrt{-3}\right)$.
Note that $[\mathbb{Q}(a):\mathbb{Q}]=2$ and $[\mathbb{Q}(b):\mathbb{Q}]=2$.
Since $a\in\mathbb{R}$, and $b\not\in\mathbb{R}$, it follows that $[\mathbb{Q}(a,b):\mathbb{Q}(a)] > 1$.
From $[\mathbb{Q}(b):\mathbb{Q}]=2$, we get $[\mathbb{Q}(a,b):\mathbb{Q}(a)]\le 2$, hence $[\mathbb{Q}(a,b):\mathbb{Q}(a)]=2$
Hence, we get $$ [K:\mathbb{Q}] = [\mathbb{Q}(a,b):\mathbb{Q}] = [\mathbb{Q}(a,b):\mathbb{Q(a)}] {\,\cdot\,} [\mathbb{Q}(a):\mathbb{Q}] = 2{\,\cdot\,}2 = 4 $$