The problem is to simplify the expression $(4x^2 - 9) \times \frac{2x+3}{2x-3}$.

1. Problem Description

The problem is to simplify the expression $(4x^2 - 9) \times \frac{2x+3}{2x-3}$.

2. Solution Steps

First, we factor the term $4x^2 - 9$. This is a difference of squares, so we can write it as:

$4x^2 - 9 = (2x)^2 - (3)^2$

$4x^2 - 9 = (2x - 3)(2x + 3)$

Now we substitute this factorization into the original expression:

$(4x^2 - 9) \times \frac{2x+3}{2x-3} = (2x - 3)(2x + 3) \times \frac{2x+3}{2x-3}$

Now we can cancel out the $(2x - 3)$ terms in the numerator and denominator, assuming $2x - 3 \ne 0$, so $x \ne \frac{3}{2}$:

$(2x - 3)(2x + 3) \times \frac{2x+3}{2x-3} = (2x+3)(2x+3)$

Finally, we can write this as:

$(2x+3)(2x+3) = (2x+3)^2$

Expanding this gives us:

$(2x+3)^2 = (2x)^2 + 2(2x)(3) + (3)^2 = 4x^2 + 12x + 9$

3. Final Answer

$4x^2 + 12x + 9$