In this specific case, you have, from your first equation:
$$ac=2$$
and, since $a,c\in\mathbb{Z}$, and $2>0$ you have the following possible pairs:
$$\begin{array}{|c|c|}
\hline
a & c\\
\hline
2 & 1\\
\hline
1 & 2\\
\hline
-1 & -2\\
\hline
-2 & -1\\
\hline
\end{array}$$
In the same way, since $b,d\in\mathbb{Z}$, we have that:
$$\begin{array}{|c|c|}
\hline
b & d\\
\hline
3 & -1\\
\hline
1 & -3\\
\hline
-1 & 3\\
\hline
-3 & 1\\
\hline
\end{array}$$
Now, due to the second equation:
$$ad+bc=-5$$
we can see that the solution is $(a,b,c,d)=(2,1,1,-3)$.
Now, in general, the problem of finding a factoriasation of $$p(x)=p_0+p_1x+p_2x^2,\ p_0,p_1,p_2\in\mathbb{Z}$$
in the form of:
$$E(Ax+B)(Cx+D)$$
with integers coefficients, boils down to finding rational roots of the equation $f(x)=0$, since, if $\frac{a}{b}$ and $\frac{c}{d}$ are the two roots of $f(x)=0$, then:
$$p(x)=p_2(x-\frac{a}{b})(x-\frac{c}{d})=\frac{p_2}{ad}(ax-b)(cx-d)$$
So, if $x=\frac{a}{b}$ is a rational root of $p(x)=0$ with $(a,b)=1$, we will have:
$$p\left(\frac{a}{b}\right)=0\Rightarrow p_0+p_1\frac{a}{b}+p_2\frac{a^2}{b^2}=0\Rightarrow b^2p_0=a(p_1b+p_2a)\Rightarrow b^2p_0|a\overset{(a,b)=1}{\Rightarrow}p_0|a$$
So, our first result is that $p_0$ should be a divisor of $a$, if $\frac{a}{b}$ is a root of $p(x)=0$. Now, we also note that - in case $a\neq0$, since this is a trivial case, due to the fact that $p_0=0$ and $p(x)=x(p_1+p_2x)$:
$$p_0+p_1\frac{a}{b}+p_2\frac{a^2}{b^2}=0\Leftrightarrow p_2+p_1\frac{b}{a}+p_0\frac{b^2}{a^2}=0$$
So, we can see that, as previously, $p_2|b$ is needed, as well.
In this way you can find several similar criteria.
Note: The conditions $p_0|a$ and $p_2|b$ are necessary conditions in order for $p$ to have rational roots, so it does not mean that $p$ has rational roots if $p_0|a$ and $p_2|b$, but that if $p$ has rational roots, then $p_0|a$ and $p_2|b$.
