A function is even if $q(-x) = q(x)$. So, we substitute $-x$ into the function:
$$q(-x)=(-x)^{2}+(-x)-3=x^{2}-x-3$$
Comparing this with the original function $q(x)=x^{2}+x-3$, we can see that $q(-x) \neq q(x)$, so the function is not even.
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