What's the difference between finding a splitting field over $\mathbb{Q}$ and over $\mathbb{Q}(\sqrt{-3})$? Say, for example, we consider the polynomial $f=x^3-2$. Then over $\mathbb{C}$ this has roots $2^{1/3}$ and $\frac{2^{1/3}(1\pm\sqrt{3}i)}{2}$. So its splitting field over $\mathbb{Q}$ is $\mathbb{Q}\big(2^{1/3}, \frac{2^{1/3}(1+\sqrt{3}i)}{2},\frac{2^{1/3}(1-\sqrt{3}i)}{2}\big) = \mathbb{Q}(2^{1/3}, \sqrt{3}i)$.
But then how does the answer change when we look at the splitting field of $f$ over $\mathbb{Q}(\sqrt{-3})$?
Since $\mathbb Q\subset\mathbb Q(\sqrt{-3})$ one has $\mathbb Q[x]\subset\mathbb Q(\sqrt{-3})[x]$. It follows that if $f(x)\in\mathbb Q[x]$ supposed ireducible have roots $\{x_1,x_2,\cdots,x_n\}$ then the splitting field over $\mathbb Q$ is $\mathbb Q(x_1,x_2,\cdots,x_n)$.
►if some $x_i\in\mathbb Q(\sqrt{-3})$ then the splitting field is the same over $\mathbb Q$ or over $\mathbb Q(\sqrt{-3})$ (What is not equal is the degree of both extensions).
►If none of the $x_i$ belongs to $\mathbb Q(\sqrt{-3})$ then both s.f. are equal and the degree of the corresponding extensions are equal too. For example the s.f. of $f(x)=x^2-5$ is generated by $\sqrt5$ over $\mathbb Q$ and over $\mathbb Q(\sqrt{-3})$ and the degree of both extensions are equal to $2$ (the number $\sqrt5$ does not belong to neither $\mathbb Q$ nor $\mathbb Q(\sqrt{-3})$.