Answer
$f(x)$ has a removable discontinuity at $x=-2$ and a non-removable discontinuity at $x=5.$
Work Step by Step
$f(x)=\dfrac{x+2}{x^2-3x-10}=\dfrac{x+2}{(x+2)(x-5)}=\dfrac{1}{x-5}\to x\ne-2$ and $x\ne5\to$
A removable discontinuity at $x=-2$ and an irremovable discontinuity (vertical asymptote) at $x=5.$
