To expand \((2 + 3x)^{-2}\), we can use the binomial series expansion for negative exponents:

\((1 + u)^{-n} = 1 - nu + \frac{n(n+1)}{2!}u^2 - \cdots\)

Let \(u = \frac{3}{2}x\) and \(n = 2\), so \((2 + 3x)^{-2} = \left(2(1 + \frac{3}{2}x)\right)^{-2} = \frac{1}{4}(1 + \frac{3}{2}x)^{-2}\).

Now expand \((1 + \frac{3}{2}x)^{-2}\):

\(1 - 2\left(\frac{3}{2}x\right) + \frac{2 \cdot 3}{2!}\left(\frac{3}{2}x\right)^2\)

\(= 1 - 3x + \frac{9}{2}x^2\)

Multiply by \(\frac{1}{4}\):

\(\frac{1}{4}(1 - 3x + \frac{9}{2}x^2) = \frac{1}{4} - \frac{3}{4}x + \frac{27}{16}x^2\)