Ta có:
$4(x^2+4xy-2y^2)-3(2x^2-xy+3y^2)=0$
$\Leftrightarrow -2x^2+19xy-17y^2=0$
$\Leftrightarrow (x-y)(2x-17y)=0$
$\Leftrightarrow \left[\begin{array}{l}x=y\\x=\dfrac{17}{2}y\end{array}\right.$
Với $x=y$ suy ra: $3y^2=3 \Leftrightarrow y=\pm1$
Với $x=\dfrac{17}{2}y$ suy ra: $\dfrac{417}{4}y^2=3 \Leftrightarrow y=\dfrac{\pm2}{\sqrt{139}}$
Suy ra: $(x;y)\in\{(1;1);(-1;-1);(\dfrac{2}{\sqrt{139}};\dfrac{17}{\sqrt{139}});(\dfrac{-2}{\sqrt{139}};\dfrac{-17}{\sqrt{139}})$