The question is inspired by: Solve the equation $3x^2+2y^2-4xy+x-2=0$ in the set of integer numbers..
The first way to solve the equation is by solving the quadratic equation for $x$ - $x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ and then we know that $x$ is an integer so the $\Delta$ has to be a perfect square, and from there you can easily get the answer. I thought of another way:
What if we calculate the min and max point of the equation and then we will take the y's integers in that range, and then calculate the x's and from there, we will take the $(x,y)$ which are integers. I thought that maybe the canonic version of the equation might help, so I calculated it:
The matrix describing the square part is $A=\begin{pmatrix}3&-2 \\ -2&2\end{pmatrix}$.
The linear vector is $J=\begin{pmatrix}1\\0\end{pmatrix}$ and the free variable is $h=-2$.
We can calculate that:
$\begin{pmatrix}3&-2 \\ -2&2\end{pmatrix}=\begin{pmatrix}\frac{-1-\sqrt{17}}{\sqrt{34+2\sqrt{17}}} & \frac{\sqrt{17}-1}{\sqrt{34-2\sqrt{17}}}\\ \frac{4}{\sqrt{34+2\sqrt{17}}}& \frac{4}{\sqrt{34-2\sqrt{17}}}\end{pmatrix}\cdot\begin{pmatrix}\frac{5+\sqrt{17}}{2}&0\\ 0&\frac{5-\sqrt{17}}{2}\end{pmatrix}\cdot{\begin{pmatrix}\frac{-1-\sqrt{17}}{\sqrt{34+2\sqrt{17}}} & \frac{\sqrt{17}-1}{\sqrt{34-2\sqrt{17}}}\\ \frac{4}{\sqrt{34+2\sqrt{17}}}& \frac{4}{\sqrt{34-2\sqrt{17}}}\end{pmatrix}}^t$
We will write that: $A=QDQ^{t}$. And, $\begin{pmatrix} u\\v \end{pmatrix}=Q\cdot\begin{pmatrix}x\\y\end{pmatrix}$.
$\begin{pmatrix}u& v\end{pmatrix}\cdot D\cdot \begin{pmatrix}u\\ v\end{pmatrix}+J^t\cdot Q\cdot \begin{pmatrix}u\\ v\end{pmatrix}+h=0\longrightarrow\frac{5+\sqrt{17}}{2}u^2+\frac{5-\sqrt{17}}{2}v^2+\begin{pmatrix} \frac{-1-\sqrt{17}}{\sqrt{34+2\sqrt{17}}}& \frac{\sqrt{17}-1}{\sqrt{34-2\sqrt{17}}}\end{pmatrix}\cdot\begin{pmatrix}u\\ v\end{pmatrix}-2=0 \\ \longrightarrow \frac{5+\sqrt{17}}{2}(u^2+\frac{\frac{-1-\sqrt{17}}{\sqrt{34+2\sqrt{17}}}}{\frac{5+\sqrt{17}}{2}}u)+\frac{5-\sqrt{17}}{2}(v^2+\frac{ \frac{\sqrt{17}-1}{\sqrt{34-2\sqrt{17}}}}{\frac{5-\sqrt{17}}{2}}v)-2=0\longrightarrow \\ \longrightarrow \frac{5+\sqrt{17}}{2}(u^2+\frac{3-\sqrt{17}}{\sqrt{34+2\sqrt{17}}}u)+\frac{5-\sqrt{17}}{2}(v^2+\frac{3+\sqrt{17}}{\sqrt{34-2\sqrt{17}}}v)-2=0\longrightarrow \\ \longrightarrow \frac{5+\sqrt{17}}{2}(u^2+\frac{3-\sqrt{17}}{\sqrt{34+2\sqrt{17}}}u+\frac{17-4\sqrt{17}}{68}-\frac{17-4\sqrt{17}}{68})+\frac{5-\sqrt{17}}{2}(v^2+\frac{3+\sqrt{17}}{\sqrt{34-2\sqrt{17}}}v+\frac{17+4\sqrt{17}}{68}-\frac{17+4\sqrt{17}}{68})-2=0\longrightarrow \\ \longrightarrow \frac{5+\sqrt{17}}{2}((u+\frac{3-\sqrt{17}}{2\sqrt{34+2\sqrt{17}}})^2-\frac{17-4\sqrt{17}}{68})+\frac{5-\sqrt{17}}{2}((v+\frac{3+\sqrt{17}}{2\sqrt{34-\sqrt{17}}})^2-\frac{17+4\sqrt{17}}{68})-2=0\longrightarrow \\ \longrightarrow \frac{5+\sqrt{17}}{2}(u+\frac{3-\sqrt{17}}{2\sqrt{34+2\sqrt{17}}})^2-\frac{17-3\sqrt{17}}{136}+\frac{5-\sqrt{17}}{2}(v+\frac{3+\sqrt{17}}{2\sqrt{34-\sqrt{17}}})^2-\frac{17+3\sqrt{17}}{136}-2=0$
Now: $\begin{pmatrix} w\\ z\end{pmatrix}=\begin{pmatrix}u+\frac{3-\sqrt{17}}{2\sqrt{34+2\sqrt{17}}}\\ v+\frac{3+\sqrt{17}}{2\sqrt{34-\sqrt{17}}} \end{pmatrix}$, and so:
$\frac{5+\sqrt{17}}{2}w^2+\frac{5-\sqrt{17}}{2}z^2=2\frac{1}{4}\Longrightarrow \frac{w^2}{\frac{9}{10+2\sqrt{17}}}+\frac{z^2}{\frac{9}{10-2\sqrt{17}}}=1$
And so we get that the canonic version of $3x^2+2y^2-4xy+x-2=0$ is $\frac{w^2}{\frac{9}{10+2\sqrt{17}}}+\frac{z^2}{\frac{9}{10-2\sqrt{17}}}=1$ in the {$z,w$} axis.
For refrence, those are the shapes in demsos:
What do I do from here?