For how many integers $x$ is $|15x^2-32x-28|$ a prime number?

$ 
\textbf{a)}\ 0
\qquad\textbf{b)}\ 1
\qquad\textbf{c)}\ 2
\qquad\textbf{d)}\ 4
\qquad\textbf{e)}\ \text{None of above}
$
1. We start with the given expression \( |15x^2 - 32x - 28| \) and need to determine for how many integer values of \( x \) this expression is a prime number.
2. First, we factorize the quadratic expression inside the absolute value:
   \[
   15x^2 - 32x - 28 = (3x + 2)(5x - 14)
   \]
3. We need \( |(3x + 2)(5x - 14)| \) to be a prime number. A prime number has exactly two distinct positive divisors: 1 and itself. Therefore, either \( (3x + 2) = 1 \) or \( (5x - 14) = 1 \), or their negatives.

4. We solve the equations:
   - \( 3x + 2 = 1 \):
     \[
     3x + 2 = 1 \implies 3x = -1 \implies x = -\frac{1}{3} \quad (\text{not an integer})
     \]
   - \( 5x - 14 = 1 \):
     \[
     5x - 14 = 1 \implies 5x = 15 \implies x = 3 \quad (\text{integer})
     \]
   - \( 3x + 2 = -1 \):
     \[
     3x + 2 = -1 \implies 3x = -3 \implies x = -1 \quad (\text{integer})
     \]
   - \( 5x - 14 = -1 \):
     \[
     5x - 14 = -1 \implies 5x = 13 \implies x = \frac{13}{5} \quad (\text{not an integer})
     \]

5. Therefore, the integer solutions for \( x \) are \( x = 3 \) and \( x = -1 \).

6. We verify that these values make the expression a prime number:
   - For \( x = 3 \):
     \[
     |(3(3) + 2)(5(3) - 14)| = |(9 + 2)(15 - 14)| = |11 \cdot 1| = 11 \quad (\text{prime})
     \]
   - For \( x = -1 \):
     \[
     |(3(-1) + 2)(5(-1) - 14)| = |(-3 + 2)(-5 - 14)| = |-1 \cdot -19| = 19 \quad (\text{prime})
     \]

7. Thus, there are exactly 2 integer values of \( x \) for which \( |15x^2 - 32x - 28| \) is a prime number.

The final answer is \( \boxed{2} \)