Find all prime $p$ such that the equation $x^2+px-330p=0$ has exactly one positive integral root. From the equation the two factors of $-330p$  add up to $p$. But I wasn't able to set up a equation based the conditions given.
 A: Note that $f(x)=(x-r)(x+p+r)=x^2+px-r(p+r)=0$ is an equation with root $r$.
This is of the form you require iff $r(p+r)=330p$
From this $p|r$ or $p|(p+r)$ - but in the second case we also have $p|r$ or $r=kp$. Hence $$kp(p+kp)=330p; kp(k+1)=330$$
Since then $p|330$ it remains to test $p=2,3,5,11$ to see what works.
I've left some stuff out for you to fill in the gaps, because you didn't give much working or context in the question.
A: I find $p = 3$. We have: $x^2 = p(330 - x)$. So $p$ divides $x^2$, and it is a prime so $p$ divides $x$. Write $x = kp$ for some natural number $k$. Then: $(kp)^2 + p(kp) - 330p = 0 $gives:
$$pk^2 + pk - 330 = 0 \Rightarrow pk(k+1) = 330$$. But $330 = 3*10*11$. So this happens only when $k = 10$, and $p = 3$. And $x = kp = 10*3 = 30$. But also that $330 = 5*6*11$. So $k = 5$, and $p = 11$ will also work. So $x = kp = 5*11 = 55$ is another answer.
A: The equation has a real positive solution and a real negative solution. If $x$ is an integer solution, then $p$ divides $x$. Let $x=py$. substituting, we find that $p^2y^2+p^2y-330p=0$, and therefore $p(y^2+y)=330$.
It follows that the only candidates for $p$ are $p=2$, $3$, $5$, or $11$. Check them all out. 
