$$
\(6 x-\frac{15}{x}>-1\)
$$


And we need to determine which of the provided options correctly describes the values of $x$ that satisfy this inequality. The options are:
(A) $\(x>-\frac{5}{3}\)$
(B) $\(-\frac{5}{2}<x<\frac{3}{2}\)$
(C) $\(\frac{-5}{2}<x<0$ and $x>\frac{3}{2}\)$
(D) $\(x<\frac{-5}{2}\)$ and $\(0<x<2\)$
(E) $\(-3<x<2\)$

Step 1: Rewrite the Inequality
First, let's bring all terms to one side to have a single inequality:

$$
\(6 x-\frac{15}{x}+1>0\)
$$


Step 2: Combine the Terms
To combine the terms, let's find a common denominator, which is $x$ :

$$
\(\frac{6 x^2-15+x}{x}>0\)
$$


So, the inequality becomes:

$$
\(\frac{6 x^2+x-15}{x}>0\)
$$

Step 3: Factor the Numerator
Now, let's factor the numerator $\(6 x^2+x-15\)$.
We're looking for two numbers that multiply to $\(6 \times(-15)=-90\)$ and add to 1 .
After some trial, the numbers are 10 and -9 :

$$
\(\begin{gathered}
6 x^2+10 x-9 x-15=0 \\
2 x(3 x+5)-3(3 x+5)=0 \\
(2 x-3)(3 x+5)=0
\end{gathered}\)
$$


So, the factored form is:

$$
\(\frac{(2 x-3)(3 x+5)}{x}>0\)
$$


Step 4: Find Critical Points
The critical points are the values of $x$ that make the numerator or denominator zero:
1. $2\( x-3=0 \Rightarrow x=\frac{3}{2}\)$
2. $\(3 x+5=0 \Rightarrow x=-\frac{5}{3}\)$
3. $x=0$ (since the denominator cannot be zero)

These critical points divide the number line into intervals where the expression's sign can be determined.

Step 5: Determine the Intervals
The critical points divide the number line into four intervals:
1. $\(x<-\frac{5}{3}\)$
2. $\(-\frac{5}{3}<x<0\)$
3. $\(0<x<\frac{3}{2}\)$
4. $\(x>\frac{3}{2}\)$

Step 6: Test Each Interval
We'll pick a test point from each interval to determine the sign of the expression $\(\frac{(2 x-3)(3 x+5)}{x}\)$.
1. Interval $\(x<-\frac{5}{3}\)$ (e.g., $\(x=-2\)$ ):
- $\(2 x-3=2(-2)-3=-7\)$ (negative)
- $\(3 x+5=3(-2)+5=-1\)$ (negative)
- $\(x=-2\)$ (negative)
$\(\circ \frac{(-)(-)}{-}= \pm=-\)$ (negative)

The expression is negative here; does not satisfy $>0$.
2. Interval $\(-\frac{5}{3}<x<0\)$ (e.g., $\(x=-1\)$ ):
$\(\circ 2 x-3=2(-1)-3=-5\)$ (negative)
- $\(3 x+5=3(-1)+5=2\)$ (positive)
- $\(x=-1\)$ (negative)


The expression is positive here; satisfies $>0$.
3. Interval $\(0<x<\frac{3}{2}\)$ (e.g., $x=1$ ):
- $\(2 x-3=2(1)-3=-1\)$ (negative)
- $\(3 x+5=3(1)+5=8\)$ (positive)
- $x=1$ (positive)
$\(\circ \frac{(-)(+)}{+}=\frac{\overline{+}}{+}=-\)$ (negative)
The expression is negative here; does not satisfy $>0$.
4. Interval $\(x>\frac{3}{2}\)$ (e.g., $x=2$ ):
- $\(2 x-3=2(2)-3=1\)$ (positive)
- $\(3 x+5=3(2)+5=11\)$ (positive)
- $x=2$ (positive)
- $\(\frac{(+)(+)}{+}=\frac{ \pm}{+}=+\)$ (positive)

The expression is positive here; satisfies > 0 .

Step 7: Consider Critical Points
The inequality is strict ( $>$ ), so the critical points where the expression equals zero ( $\(x=-\frac{5}{3}\)$ and $\(x=\frac{3}{2}\)$ ) are not included in the solution. Also, $x=0$ is excluded because the expression is undefined there.

Step 8: Combine the Valid Intervals
From our testing, the expression is positive in:
- $\(-\frac{5}{3}<x<0\)$
- $\(x>\frac{3}{2}\)$

Now, let's see which option matches this.

Step 9: Compare with Given Options
Let's look at the options:
(A) $\(x>-\frac{5}{3}\)$
- This includes $\(x>\frac{3}{2}\)$ but also $\(0<x<\frac{3}{2}\)$, which is incorrect. Not matching.
(B) $\(-\frac{5}{2}<x<\frac{3}{2}\)$
- Our solution includes $x>\frac{3}{2}$, which this excludes. Also, $-\frac{5}{2}$ is not a critical point we have. Not matching.
(C) $\(\frac{-5}{2}<x<0\)$ and $\(x>\frac{3}{2}\)$
- Close, but our lower bound is $\(-\frac{5}{3}\)$, not $\(-\frac{5}{2}\)$. Not matching exactly.
(D) $\(x<\frac{-5}{2}\)$ and $\(0<x<2\)$
- Our solution doesn't include $\(x<-\frac{5}{3}\)$ (it's negative there), and $\(0<x<\frac{3}{2}\)$ is negative. Not matching.
(E) $\(-3<x<2\)$
- This is too broad and includes regions where the inequality doesn't hold. Not matching.

Wait a minute, none of the options exactly match $\(-\frac{5}{3}<x<0\)$ and $\(x>\frac{3}{2}\)$.
But let's look at option (C) again: $\(\frac{-5}{2}<x<0\)$ and $\(x>\frac{3}{2}\)$.
Our solution is $\(-\frac{5}{3}<x<0\)$ and $\(x>\frac{3}{2}\)$.
Now, $\(-\frac{5}{3} \approx-1.666 \ldots\)$ and $\(-\frac{5}{2}=-2.5\)$.
So, $\(\frac{-5}{2}<x<0\)$ includes $\(-\frac{5}{3}<x<0\)$, plus $\(-\frac{5}{2}<x<-\frac{5}{3}\)$.
In $\(-\frac{5}{2}<x<-\frac{5}{3}\)$, the expression is negative (from our first interval test), so the inequality doesn't hold there.

Thus, option (C) includes some $x$ where the inequality doesn't hold, so it's not entirely correct.