We only find interger roots of this equation.
\(2x^2-5xy+3y^2=7\)
\(\Leftrightarrow\left(x-y\right)\left(2x-3y\right)=7\)
Case 1: \(\left\{{}\begin{matrix}x-y=1\\2x-3y=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-5\end{matrix}\right.\)
Case 2: \(\left\{{}\begin{matrix}x-y=7\\2x-3y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=20\\y=13\end{matrix}\right.\)
Case 3: \(\left\{{}\begin{matrix}x-y=-1\\2x-3y=-7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\)
Case 4: \(\left\{{}\begin{matrix}x-y=-7\\2x-3y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-13\end{matrix}\right.\)
