If $ x,y ∈\Bbb{Z} $ find $x$ and $y$ given: $2x^2-3xy-2y^2=7$ We are given an equation: 
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y ∈\Bbb{Z}$.
After we subtract 7 from both sides, it's clear that this is quadratic equation in its standard form, where $a$ coefficient equals to 2, $b=-3y$ and $c=-2y^2-7$. Thus discriminant equals to $9y^2+16y^2+56=25y^2+56$.
Now $x = \frac{3y±\sqrt{25y^2+56}}{4}$ , $x$ to be an integer $\sqrt{25y^2+56}$ has to be an integer too. I substituted $y$ as nonnegative integer first, because the answer wouldn't differ anyway as it is squared and found that when $y=1$, $\sqrt{25y^2+56}=9$, so we get $x = \frac{3±9}{4}$ and when we have plus sign we get $x = \frac{3+9}{4}=3$. So there we have it, $x=3, y=1$ is on of the solution. But $y=-1$ will also work because $y$ is squared, again, we get $x = \frac{-3±9}{4}$, if we have minus sign we have $x=-3$. Thus another solution, leaving us with:
$$ x=3, y=1$$
$$ x=-3, y=-1$$
I checked other integers for $y$ but none of them lead to solution where $x$ is also an integer. But here's problem, how do I know for sure that these two solutions are the only solutions, I can't obviously keep substituting $y$ as integers, as there are infinitely many integers. So that's why I came here for help.

Hint:
$$2x^2-3xy-2y^2=2x^2-4xy+xy-2y^2=2x\underbrace{(x-2y)}+y\underbrace{(x-2y)}=?$$
