Answer
$ x=2$ or $x=\pm\sqrt{5}$
Work Step by Step
We factor by grouping and solve:
$x^{3}-2x^{2}-5x+10=0$
$x^{2}(x-2)-5(x-2)=0$
$(x-2)(x^{2}-5)=0$
$x-2=0$ or $x^2-5=0$
$ x=2$ or $x=\pm\sqrt{5}$
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