For example, solve this : $x^2 +2x +3 = 10y^2+11y+12$ with x,y integers
I can reduce this equation into this : $Y^2-40X^2=-279$ with $X=x+1$ and $Y=20y+11$
Now I need to find the primitive solutions of this equation but i don't know how to do it. And these primitive solutions that make it possible to find general solutions must be in a very specific form. Here, $Y=11 $ mod $ 20$. Any help would be appreciated.
EDIT : I did a mistake in the basic equation. I changed it.
This is not a complete answer,
but I did some calculations and found the following solutions to $Y^2-40X^2=-279: $
$(Y,X)=$ $ (9,3), (19,4), (41,7), (61, 10), (119, 19), $ and $(189,30)$.
If $(Y,X)$ is a solution, then so are $(-Y,X), (Y,-X), $ $(-Y,-X),$
and $(19Y+120X,3Y+19X)$. [Note: $19^2-40\cdot3^2=1$.]
Then we need to take solutions where $Y=20y+11$,
such as $(-9,\pm3), (-189,\pm30), $ and $(531,\pm84)$, which yield solutions $(y,x)=$
$(-1,-4), (-1, 2), (-10,-31), (-10, 29), (26, -85), $ and $(26,83)$ to your original equation.