$n=3$ or $n=-4$
Work Step by Step
$n^2+n-12=0$ $n^2-3n+4n-12=0$ $n(n-3)+4(n-3)=0$ $(n-3)(n+4)=0$ $n=3$ or $n=-4$ The equation has two solutions, $n=3$ or $n=-4$.
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