We can do this by substituting different values of $x$ into the equation and solving for $y$.
For $x=-2$, we have $y=(-2)^{2}+2=4+2=6$. So, $(-2,6)$ is a solution.
For $x=-1$, we have $y=(-1)^{2}+2=1+2=3$. So, $(-1,3)$ is a solution.
For $x=2$, we have
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