\(3\left(x+1\right)^2=-4y^2-3y+7\)
\(\Rightarrow-4y^2-3y+7\ge0\Rightarrow-\frac{7}{4}\le y\le1\)
\(\Rightarrow y=\left\{-1;0;1\right\}\)
- Với \(y=-1\Rightarrow3\left(x+1\right)^2=6\Rightarrow\) ko có x nguyên t/m
- Với \(y=0\Rightarrow3\left(x+1\right)^2=7\) ko có x nguyên t/m
- Với \(y=1\Rightarrow3\left(x+1\right)^2=0\Rightarrow x=-1\)