Casebash's answer is very good.
Here is a second answer. You can apply the following
Theorem: If the roots $x_{1},x_{2}$ of $f(x)=ax^{2}+bx+c$ are real and $x_{1}\neq x_{2}$ (with $x_{1} < x_{2}$), then, the signal of $f(x)$ is:
- opposite to the signal of $a$ for $x\in \left[ x_{1},x_{2}\right] $,
- the same of $a$ for $x\in \left] -\infty ,x_{1}\right[ \vee x\in \left] x_{2},-\infty \right[ $.
Since in your case $a=1>0$, $x_{1}=-2<5=x_{2}$, you have $x^{2}-3x-10>0$ for $x\in \left] -\infty ,-2\right[ \vee x\in \left] 5,\infty \right[ $.
Addendum: A possible proof of this theorem is to use the explanation of Casebash, taking into consideration that $ax^{2}+bx+c=a(x-x_1)(x-x_2)$