A.P. terms in a Quadratic equation. 
The terms $a,b,c$ of quadratic equation $ax^{2}+bx+c=0$ are in A.P. and positive. Let this equation have integral root $\alpha,\ \beta$. Then find the value of $\alpha+ \beta + \alpha \cdot \beta$ ?

please point where I'm wrong:
Let common difference be $d$
$\implies \alpha+ \beta + \alpha \cdot \beta=\dfrac{c-b}{a}=\dfrac{d}{a} \implies a|d \ \ \ \ \longrightarrow  \ \ \ \ \              \because (b=a+d$, $c=a+2d)$
Also
, $ax^{2}+(a+d)x+(a+2d)=0$. 
$\implies$ $\alpha,\ \beta=\dfrac{-(a+d) \pm \sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}}{2a}$. 
For this to be integer $\sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}$ must be perfect square.
$\implies$ ${(a+d)^{2}-4\cdot a \cdot (a+2d)}=p^{2}$ for some $p$.
$\implies -3a^{2}+d^{2}-6ad=p^{2}$
$\implies -3a^{2}+a^{2}q^{2}-6a^{2}q=p^{2}$ $\because$ $a|d \implies aq=d$ for some $q$.
$\implies a^{2}(-3+q^{2}-6q)=p^{2}$ 
$\implies -3+q^{2}-6q\ $  has to be perfect square. By trial $q=7$
But I need to get this without trial, please help.

We have, $$p(x)=ax^2+bx+c=ax^2+(a+p)x+(a+2p)$$
and thus, $$t+r+r\cdot t=-\frac{a+p}{a}+\frac{a+2p}{a}=\frac{p}{a}$$
then $p=a\cdot k$, $k=(t+r+t\cdot r) \in \Bbb Z$.
Then our polynomial becomes,$$p(x)=ax^2+a(1+k)x+a(1+2k)$$
$$t=\frac{-(1+k) \pm \sqrt{k^2-6k-3}}{2} \quad...(1)$$
So 
$$k^2-6k-3=q^2 \Rightarrow (k-3)^2-12=q^2 \Rightarrow (k+q-3)(k-q-3)=12$$
and split $12$ as a product of two integer and find all possible values for $k$.
Take for example:
\begin{cases}
k+q-3=6 \Rightarrow k+q=9 \\
k-q-3=2 \Rightarrow k-q=5
\end{cases}
Adding up both equations we get $2k=14 \Rightarrow k=7$
And back to $(1)$, we get $(t,r)=(-3,-5)$ or $(t,r)=(-5,-3)$.
