How many $ax^2+bx+c=0$ with distinct pairs of rational roots can be made from integers $N=|ac|=|\alpha\beta|$ and $b=\alpha+\beta$? Only for illustration, let's choose an integer $N=6$. I want to create a collection of all possible quadratic equations in the form of $ax^2+bx+c=0$ where

*

*$|ac|=|\alpha\beta|=N=6$, and

*$\alpha+\beta=b$.

Note that all constants are integers but $x$ is not necessarily integer.
Here is my attempt.
How many ways to specify $a$ and $c$
The prime factors of $N$ is
$$
P=\{2,3\}
$$
The all possible pairs of $(a,c)$ are
$$
\{(1,\pm 6), (6,\pm 1), (2,\pm 3), (3,\pm 2)\}
$$
Here I don't need to consider the case in which $a<0$ because, for example,
$6x^2+5x-1=0$ is actually identical to $-6x^2-5x+1=0$. Only the sign of $b$ and $c$ do matter here.
From $2^2=4$ subsets of $P$, we can create $\frac{2^2}{2}=2$ "partitions", each "partition" can be assigned to $(a,c)$ in $2!=2$ ways, and $c$ can have two choice of signs. Thus there are
$$
\frac{2^2}{2}\times 2! \times 2 = 8
$$
ways to create $(a,c)$.
How many ways to specify $b$
The all possible pairs of $(\alpha,\beta)$ are
$$
\{(\pm 1,\pm 6), (\pm 2,\pm 3)\}
$$
Therefore  the all possible values of $b$ are
$$
\{\pm |1+6|, \pm |1-6| = \pm |2+3|, \pm |2-3|\} = \{\pm 7, \pm 5, \pm 1 \}
$$
I cannot find an easier way to calculate how many possible ways to determine $b$ because the same $b$ can be obtained from two (or possibly more) "partitions". For example, partition $\{1,6\}$ and $\{2,3\}$ can produce $b=\pm 5$ as follows.
$$
\pm |1-6| = \pm |2+3|
$$
As we can see, there are $3\times 2=6$ ways to assign $b$.
Final
Thus in total, we have $8\times 6=48$ quadratic equations. I have not checked programmatically whether all of these equations have distinct pair of roots.
Question
Generally speaking, for any positive integer $N$, how many quadratic equations (with the constraints given above) are possible to make?
 A: This is a partial answer.
This answer deals with $N$ such that $N\not\equiv 0\pmod 3$.
This answer proves that the number of such quadratic equations is
$$\begin{cases}2\sigma_0(N)^2&\text{if $N$ is not a square number with $N\not\equiv 0\pmod 3$}
\\\\\sigma_0(N)(2\sigma_0(N)-1)&\text{if $N$ is a square number with $N\equiv 1\pmod 3$}\end{cases}$$
where $\sigma_0(N)$ is the number of the positive divisors of $N$.
Proof :
We may suppose that $a\gt 0$.
Note that $ax^2+bx+c=0$ has two distinct rational roots if and only if $b^2-4ac$ is a non-zero square number.
Also, let $\sigma_0(N)$ be the number of the positive divisors of $N$.
($\sigma_0(N)=\displaystyle\prod_{k=1}^{d}(e_k+1)$ when $N=\displaystyle\prod_{k=1}^{d}p_k^{e_k}$ where $p_1,p_2,\cdots,p_k$ are distinct prime numbers.)

*

*Case 1 : $ac=\alpha\beta$Since we have $b^2-4ac=(\alpha+\beta)^2-4\alpha\beta=(\alpha-\beta)^2$, we see that $ax^2+bx+c=0$ has two distinct rational roots if and only if $\alpha\not=\beta$.If $\alpha\beta=\alpha'\beta',\alpha'\not=\alpha$ and $\alpha'\not=\beta$, then $$\small\alpha+\beta-(\alpha'+\beta')=\alpha+\beta-\alpha'-\frac{\alpha\beta}{\alpha'}=\frac{(\alpha-\alpha')(\alpha'-\beta)}{\alpha'}\not=0\implies \alpha+\beta\not=\alpha'+\beta'\tag1$$Case 1-1 : $ac=\alpha\beta=N$The number of $(a,c)$ is given by $\sigma_0(N)$ since $a$ is positive.If $N$ is not a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.If $N$ is a square number, then the number of $(\alpha,\beta)$ is given by $2(\sigma_0(N)-1)$ since the case where $\alpha=\beta$ has to be excluded. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)-1$. So, the number of such quadratic equations is $\sigma_0(N)(\sigma_0(N)-1)$.Case 1-2 : $ac=\alpha\beta=-N$The number of $(a,c)$ is given by $\sigma_0(N)$ since $a$ is positive.If $N$ is not a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.If $N$ is a square number, then the number of $(\alpha,\beta)$ is given by $2\sigma_0(N)$ since the case where $\alpha=\beta$ does not happen. Since $b=\alpha+\beta$ is symmetric, it follows from $(1)$ that the number of $b$ is given by $\sigma_0(N)$. So, the number of such quadratic equations is $\sigma_0(N)^2$.


*Case 2 : $ac=-\alpha\beta=\pm N$For $N$ such that $N\not\equiv 0\pmod 3$, we have
$$\begin{align}-\alpha\beta=\pm N&\implies (\alpha,\beta)\equiv (1,1),(1,2),(2,1),(2,2)\pmod 3
\\\\&\implies b^2-4ac=(\alpha+\beta)^2+4\alpha\beta\equiv 2\pmod 3\end{align}$$So, $b^2-4ac$ cannot be a square number.
Hence, it follows from the above cases that the number of such quadratic equations is
$$\begin{cases}2\sigma_0(N)^2&\text{if $N$ is not a square number with $N\not\equiv 0\pmod 3$}
\\\\\sigma_0(N)(2\sigma_0(N)-1)&\text{if $N$ is a square number with $N\equiv 1\pmod 3$}\end{cases}$$
where $\sigma_0(N)$ is the number of the positive divisors of $N$.
