What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$

This is almost correct, but I believe the original sum needs to range from $-\infty$ to $\infty$ instead of $0$ to $\infty$. The solution that follows considers the sum $\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}$, and throughout I will write $\sum_{n=-\infty}^\infty f(n)$ to mean $\lim_{N\rightarrow \infty}\sum_{n=-N}^N f(n)$.

Factoring the quadratic, with your definition of $z_{0},\ z_{1}$, we have $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}=\frac{1}{a}\sum_{n=-\infty}^{\infty}\frac{1}{\left(n-z_{0}\right)\left(n-z_{1}\right)}.$$ Assume that neither $z_0$ nor $z_1$ are integers, since otherwise we would have a $\frac{1}{0}$ term appearing in the sum. By applying partial fractions, remembering that $z_{0}-z_{1}=\frac{\sqrt{b^{2}-4ac}}{a}$ we get $$\frac{1}{\sqrt{b^{2}-4ac}}\sum_{n=-\infty}^{\infty}\left(\frac{1}{n-z_{0}}-\frac{1}{n-z_{1}}\right).$$ By the cotangent identity $\pi\cot\left(\pi x\right)=\sum_{n=-\infty}^{\infty}\frac{1}{n+x},$ we conclude that $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}=\frac{\pi\cot\left(\pi z_{1}\right)-\pi\cot\left(\pi z_{0}\right)}{\sqrt{b^{2}-4ac}}.$$

Take $a=1,\ b=3, \ c=2$, then $z_0=-2, \ z_1=-1$, and so you have to compute $\cot(-\pi)$ and $\cot(-2\pi)$ which make no sense. However $$ \sum_{n=0}^\infty\frac{1}{n^2+3n+2}=\sum_{n=0}^\infty(\frac{1}{n+1}-\frac{1}{n+2}) =\lim_{m\to \infty}(1-\frac{1}{m+2})=1. $$

Here is my answer in terms of the digamma function (cf. Abramowitz and Stegun). Using the decomposed version of the sum given by Eric Naslund♦, we find:

$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}= \frac{1}{\sqrt{b^{2}-4ac}}\sum_{n=0}^{\infty}\left(\frac{1}{n-z_{0}}-\frac{1}{n-z_{1}}\right)= \frac{\psi (-z_0)-\psi(-z_1)}{\sqrt{b^{2}-4ac}} $$

I note that this answer is in fact quite similar to the sum's closed form from $-\infty$ to $\infty$.

Here is an alternate solution:

$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}= \frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)} $$

If $c_n$ is the $n^{th}$ term of the second version, then

$$\frac{c_{n+1}}{c_n}=\frac{(n-z_0)(n-z_1)}{(n-z_0+1)(n-z_1+1)}$$

which shows that this sum can be written as a hypergeometric function

$$ \frac{1}{a}\sum_{n=0}^\infty \frac{1}{(n-z_0)(n-z_1)}= \frac{1}{az_0z_1}\sum_{n=0}^\infty \frac{\Gamma(n-z_0)\Gamma(n-z_1)n!}{\Gamma(n-z_0+1)\Gamma(n-z_1+1)n!}= \frac{1}{z_0z_1a}{_3}F_2(-z_0, -z_1, 1;1-z_0, 1-z_1;1) $$