The diophantine equation $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ can be solved completely : for every sixtupel $(A,B,C,D,E,F)$ we can determine the complete set of integer pairs satisfying the equation.
What about the more complicated diophantine equation $$Ax^2+By^2+Cz^2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0$$ ?
In short, a diophantine polynomial equation with degree $2$ in $3$ unknowns. I don't think that we can fully solve such an equation in general, can we ? In which cases can we solve it completely ?
There are so many cases that it would be hard to list all cases in which we can solve it completely. For example, all of the two-variable quadratic Diophantine equations are special cases, and there are long papers written about various cases of two-variable quadratic Diophantine equations. (For example, John Robertson's page, http://www.jpr2718.org/, has some very nice papers on solving second-degree Diophantine equations in two variables.)
I think that the general answer for the three-variable case is, for most choices of coefficients, there are infinitely many solutions, but there is no neat tidy formula that describes all such solutions. (Of course, in certain special cases, there would be.)
Some simple cases are mentioned here: http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html
I would recommend starting with the simple case $ax^2 + by^2 + cz^2 = d$, since you can usually reduce to this case by completing the square three times. The "parabolic" case $ax^2 + by^2 + cz = d$ should be solvable using modular arithmetic (mod $c$).