Answer
The domain is $(-\infty,-\frac{2}{5}] \cup [\frac{1}{3},\infty)$ .
Work Step by Step
$$
f(x)=\sqrt {15x^{2}+x-2}
$$
the values $x$ for $f(x)$ is defined when:
$$
15x^{2}+x-2 \geq 0
$$
we can find all zeros of $f(x)$ :
$$
15x^{2}+x-2 = 0
$$
$ \Rightarrow $
$$
\:x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:15\left(-2\right)}}{2\cdot \:15}
$$
$ \Rightarrow $
$$
x_{1}=\frac{1}{3},\:x_{2}=-\frac{2}{5}
$$
this numbers form the intervals $ (-\infty,-\frac{2}{5}), (-\frac{2}{5} ,\frac{1}{3}) $ and $(\frac{1}{3},\infty)$.
we observe that only values in the intervals $(-\infty,-\frac{2}{5}) $ and $(\frac{1}{3},\infty)$ satisfy the inequality.
So the domain is $(-\infty,-\frac{2}{5}] \cup [\frac{1}{3},\infty)$ .
