Answer
$\color{blue}{(-\infty, -\frac{3}{2}) \cup (-\frac{3}{2}, 5) \cup (5, +\infty)}$
Work Step by Step
The denominator of a rational expression is not allowed to be equal to zero as it will make the expression undefined.
Find the values of $x$ for which the denominator is equal to zero by equating the denominator to zero:
$(2x+3)(x-5)= 0$
Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain:
\begin{array}{ccc}
&2x+3 = 0 &\text{ or } &x-5=0
\\&2x=-3 &\text{or} &x=5
\\&x=-\frac{3}{2} &\text{or} &x=5
\end{array}
This means that the value of $x$ can be any real number except $-\frac{3}{2}$ and $5$.
Therefore, the domain of the given rational expression is the set of real numbers except $-\frac{3}{2}$ and $5$.
In interval notation, the domain is:
$\color{blue}{(-\infty, -\frac{3}{2}) \cup (-\frac{3}{2}, 5) \cup (5, +\infty)}$
