How can I solve this question on Quadratic Equations and Arithmetic Progression? Let $P (x) = ax^2 + bx + c$ where $a,b,c$ are in arithmetic progression and are positive constants. Let $P (x) = 0$ for $t$ and $r$, where  $t$ and $r$ are integers. Find $t + r +tr$.
Answer is : $7$
I tried it as follows :
$t + r = -b/a$ ;
$tr = c/a$ ;
$t + r + tr = -b/a + c/a = (c-b)/a = (b-a)/a= b/a - 1$  
Therefore,
$2t + 2r + rt = -1$
I could not figure out what to do next. Please help.

$$p(x)=ax^2+bx+c=ax^2+(a+p)x+(a+2p)$$
$$t+r+r\cdot t=-\frac{a+p}{a}+\frac{a+2p}{a}=\frac{p}{a}$$
then $p=a\cdot k$, $k=(t+r+t\cdot r) \in \Bbb Z$.
$$p(x)=ax^2+a(1+k)x+a(1+2k)$$
$$t=\frac{-(1+k) \pm \sqrt{k^2-6k-3}}{2} \quad (1)$$
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 $k$.
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 $2k=14 \Rightarrow k=7$
Backing to $(1)$ we get $(t,r)=(-3,-5)$ or $(t,r)=(-5,-3)$.
