Answer
$\dfrac{x^2+3x-15}{x-2}$
$(-\infty,2)\cup(2,\infty)$
Work Step by Step
We are given the functions:
$$\begin{cases}
f(x)=x^2+3x-15\\
g(x)=x-2.
\end{cases}$$
We write the function $\dfrac{f}{g}$:
$$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}=\dfrac{x^2+3x-15}{x-2}.$$
The function $\dfrac{f}{g}$ is defined for all real value of $x$ for which $g(x)\not=0$, therefore its domain is:
$$(-\infty,2)\cup(2,\infty).$$
