This is a homework, but I've generalized it as possible in order not to have exact answer rather that to understand the very principle of solution.
The problem is following: consider $\mathbb{K}$ a field and $E$ an extension field of $\mathbb{K}$. For a given irreducible polynomial $P(x)$ from the ring $\mathbb K[x]$ find the number of homomorphisms from the stem field for $P(x)$ to the field $E$.
A stem field for an irreducible polynomial $P$ in $\mathbb{K}[x]$ is a pair $(F,\alpha)$, where $\alpha$ is a root of $P$ and $F$ is an extension of $\mathbb{K}$, i.e. $F = \mathbb{K}[\alpha]$ and $P(\alpha)$=0
My understanding is following:
Any stem field $F$ is isomorphic to $ \dfrac{\mathbb K[x]}{(P(x))}$
The number of homomorphisms from $\dfrac{\mathbb K[x]}{(P(x))}$ to $E$ is equal to the number of roots of this particular polynomial in $E$.
Example: if $\mathbb{K} = \mathbb{Q}$ and $P(x$) has $n$ roots in $\mathbb R$ (real roots) and $m$ complex (strictly non-real) roots, then the number of homomorphisms to $\mathbb R$ is n and number of homomorphisms to $\mathbb C$ is n+m.
Is my understanding correct at all? If not, can you give me a hint in what direction I should look for.
Your understanding is correct, except for one important assumption you've left out. What you say is correct provided that $\mathbb{K}$ is a subfield of $E$ and you are only considering homomorphisms $F\to E$ which are the identity of $\mathbb{K}$. In general, to define a homomorphism $\mathbb{K}[x]/(P(x))\to E$, you have to first choose a homomorphism $\varphi:\mathbb{K}\to E$, and then choose an element of $E$ which is a root of the polynomial obtained by applying $\varphi$ to the coefficients of $P$. When you require $\varphi$ to be the inclusion map, this means your only choice is an element of $E$ which is a root of $P$. But if you do not assume this, there may be many more homomorphisms $\mathbb{K}[x]/(P(x))\to E$ that do something else on $\mathbb{K}$.