Quadratic Diophantine Equation$(x^2+x)(y^2-1)=240$ For whole numbers $x$ and $y$,  $$x,y | (x^2+x)(y^2-1)= 240$$
Find the biggest and smallest value for $x-y$.
How do you proceed with such a question?
Are their formulas or something for that type of equation?
I'd appreciate any help.

$(x^2+x)(y^2-1)=240$ implies that 
$y = \sqrt{1 + \dfrac{240}{x(x+1)}}$
Clearly both $x$ and $x+1$ need to be divisors of $240$.
These are the divisors of $240$
\begin{align}
   1 &\quad 240 \\
   2 &\quad 120 \\
   3 &\quad 80 \\
   4 &\quad 60 \\
   5 &\quad 48 \\
   6 &\quad 40 \\
   8 &\quad 30 \\
   10 &\quad 24 \\
   12 &\quad 20 \\
   15 &\quad 16
\end{align}
For $x = 1,2,3,4,5,15$ we get 
$y = 11, \sqrt{41}, \sqrt{21}, \sqrt{13}, 3, \sqrt 2$
So the minimum value of $x-y$ is $1-11 = -10$
and the maximum value of $x-y$ is $5-3 = 2$
