I heard that the discriminant of an extension is equal to the product of local discriminants. To better understand this I tried it out on some quadratic number fields, but ended up getting stuck.
For example I considered the number field $\mathbb{Q}(\sqrt{5})$. Since $$disc(\mathbb{Q}(\sqrt{5}))=-20$$ we have that the only ramified primes are 2 and 5 and so we should have $$-20=disc(\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2)\cdot disc(\mathbb{Q}_5(\sqrt{5})/\mathbb{Q}_5).$$ (I suppose knowing the discriminant before hand is kind of cheating...)
Anyway, following the answer here (and also with the help of Gouvea's book p-adic Numbers) I was able to show that $\mathbb{Q}_5(\sqrt{5})/\mathbb{Q}_5$ is ramified and thus divides $disc(\mathbb{Q}(\sqrt{5}))$. I am not sure about what to do about $disc(\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2)$. I do know that $\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2$ is unramified, but this is part of my confusion (I guess things get weird in characteristic 2). In particular I am curios as to how $disc(\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2)$ can be computed? Also, can this be done without finding an explicit basis for the ring of integers of $\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2$?
Maybe better for a full-fledged answer than a comment:
You have it wrong: though $\text{disc}(\Bbb Q(\sqrt{-5\,})=-20$, the fact is that $\text{disc}(\Bbb Q(\sqrt{5\,})=5$. This alone may dissipate all your confusions. But let me go into more detail.
Over $\Bbb Z$, an integral basis for (the integers of) $\Bbb Q(\sqrt{5\,})$ is $\{1,\frac{1+\sqrt5}2\}$. Since the number $\frac{1+\sqrt5}2$ has for its minimal polynomial $X^2-X-1$, you readily calculate the (global) discriminant to be $5$. I don’t know what general definition of the discriminant of a ring $R\supset\Bbb Z$ you’re using, but the most primitive one is to take your integral basis $\{a_1,\cdots,a_n\}$ and then calculate the determinant of the matrix whose $(i,j)$-entry is the trace of $a_ia_j$, the “trace” being the field-theoretic trace from the fraction-field of $R$ down to $\Bbb Q$.
To go locally, you look, as you did, at $\Bbb Q_5(\sqrt{5\,})\supset\Bbb Q_5$, for which a good basis is $\{1,\sqrt5\}$ or the global one I mentioned above. The discriminant ideal here is $(5)$. And at the prime $2$, you look at $\Bbb Q_2(\sqrt{5\,})\supset\Bbb Q_2$ and you have to use the global basis above; $\{1,\sqrt5\}$ is not an integral basis, even at $2$. And the discriminant ideal is $(1)\,$: that is, the extension is unramified at $2$.
If you have further questions, I can edit this answer or respond in a comment.