What is the dimension of $\mathbb{Q(\sqrt{5},i)}$ over $\mathbb{Q}?$
Considering that $\mathbb{Q} \subseteq \mathbb{Q(\sqrt{5})} \subseteq \mathbb{Q(\sqrt{5},i)}$
$[\mathbb{Q(\sqrt{5},i)} : \mathbb{Q}] = [\mathbb{Q(\sqrt{5})} : \mathbb{Q}] [\mathbb{Q(\sqrt{5},i)} : \mathbb{Q(\sqrt{5})}]$
$[\mathbb{Q(\sqrt{5}):\mathbb{Q}] = 2}$
the minimal polynomial is $f(x) = x^2-5$ in fact $f(\sqrt{5}) = 0$ and $f(x)$ has no rational roots
$[\mathbb{Q(\sqrt{5},i):\mathbb{Q(\sqrt{5})}] = 2}$
the minimal polynomial is $f(x) = x^2 +1$ in fact $f(i) = 0$ and $f(x)$ has no roots in $\mathbb{Q(\sqrt{5})}$
So $[\mathbb{Q(\sqrt{5},i)} : \mathbb{Q}] = 4$
It's right?
Final step could be more rigurous. $$[\mathbb{Q(\sqrt{5},i)}:\mathbb{Q(\sqrt{5})}] \leq 2,$$ since $t^2+1$ has coefficents in $\mathbb{Q}(\sqrt{5})$, and has $i$ as root. In the other hand $$[\mathbb{Q(\sqrt{5},i)}:\mathbb{Q(\sqrt{5})}]>1, $$ since $i\in\mathbb{C}$ and $\mathbb{Q(\sqrt{5})}\subset\mathbb{R}$. Hence the degree is $2$.