The problem states to find the max ideals of the ring $\mathbb{Z}[x] / (6,x^2+7x+12)$. It is easy to show that you only have to find the max ideals of $\mathbb{Z}_6[x] / (x^2+x)$.
In the solution it states that the max ideals of the latest ring are $J/(x^2+x)$, with $J$ an ideal of $\mathbb{Z}_6[x]$ and $x^2+x\in J$. (why?)
And since
$$ (\mathbb{Z}_6[x] / (x^2+x)) / (J / (x^2+x)) \cong \mathbb{Z}_6[x] / J $$ is a field (why?) he have that $J$ is max in $\mathbb{Z}_6[x] $.
I don’t really understand why it is a field and why a max ideal must have that form.
For any ideal $I$ in a ring $R$, isomorphism theorems for rings give a bijection between ideals of $R$ containing $I$, and those of $R/I$. This bijection is via the map $J \mapsto J/I$, and the correspondence preserves inclusion. This implies that $J$ containing $I$ is maximal in $R$ if and only if $J/I$ is maximal in $R/I$. So your problem boils down to determining the maximal ideals of $\mathbb Z_6[x]$ which contain $x^2+x$.
Taking $R = \mathbb Z_6[x]$, t's not true that any ideal $J$ which contains $x^2+x$ is maximal in $R$ (and hence not every $J/\langle x^2+x \rangle$ gives a maximal ideal in the quotient, by the above discussion). For instance, the ideal $J=\langle x^2+x \rangle $ itself is not maximal.
The part about the quotient being a field follows from the fact that an ideal $\mathfrak m \subset R$ is maximal if and only if $R/\mathfrak m$ is a field (for $R$ commutative with $1$). See here.