The inequality given is:

$2(x - 4)(x + 4) > 3x(2x + 3)$

Let's solve this inequality step by step:

1. Expand both sides:

   Left side:
   $2(x - 4)(x + 4) = 2(x^2 - 16) = 2x^2 - 32$

   Right side:
   $3x(2x + 3) = 6x^2 + 9x$

2. Rewrite the inequality:

   $2x^2 - 32 > 6x^2 + 9x$

3. Move all terms to one side:

   $2x^2 - 6x^2 - 9x - 32 > 0$

   Simplify:

   $-4x^2 - 9x - 32 > 0$

   To make it more conventional, multiply by $-1$:

   $4x^2 + 9x + 32 < 0$

4. Analyze the quadratic inequality:

   For a quadratic $ax^2 + bx + c < 0$, we need to find the roots using the discriminant:

   Discriminant ($\Delta$) is given by:

   $\Delta = b^2 - 4ac = 9^2 - 4(4)(32) = 81 - 512 = -431$

Since the discriminant is negative ($\Delta < 0$), the quadratic equation has no real roots and is always positive since $a > 0$ for $4x^2 + 9x + 32$.

Therefore, $4x^2 + 9x + 32 < 0$ has no real solutions.

Conclusion: The inequality $2(x - 4)(x + 4) > 3x(2x + 3)$ holds for no real values of $x$.

Would you like further details or have any specific questions?

Here are some related questions you might consider:

1. How can inequalities be solved graphically?
2. What are the implications of a negative discriminant in a quadratic equation?
3. How does one solve inequalities involving higher-order polynomials?
4. What techniques can be used to analyze the behavior of quadratic functions?
5. How do you determine the intervals of positivity and negativity of a quadratic function?

Tip: When solving inequalities, always check the nature of the quadratic (positive or negative discriminant) to determine if the expression can ever be less than or greater than zero.