equation

3y^2-(2x)y+(7x^2-27)=0 \\ \sqrt{D}=2\sqrt{-20x^2+81} \\ \\ y=\frac{x+\sqrt{-20x^2+81}}{3} \\ \\ y=\frac{x-\sqrt{-20x^2+81}}{3}\\ \\\texttt{now we want}\sqrt{-20x^2+81}, and \sqrt{-20x^2+81}\texttt{as perfect squares } \\\texttt{the values that satisfy are} \\ x=0,x=2,x=-2 \\ \texttt{we can have corresponding y's for these x's,hence 3 solutions}