We start by rewriting \(\frac{1}{\sqrt{4 + 3x}}\) as \((4 + 3x)^{-\frac{1}{2}}\).

Using the binomial expansion for \((1 + u)^n\), where \(u = \frac{3}{4}x\) and \(n = -\frac{1}{2}\), we have:

\((1 + \frac{3}{4}x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(\frac{3}{4}x\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(\frac{3}{4}x\right)^2 + \cdots\)

Calculating each term:

First term: \(1\)

Second term: \(-\frac{1}{2} \times \frac{3}{4}x = -\frac{3}{8}x\)

Third term: \(\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2} \times \left(\frac{3}{4}x\right)^2 = \frac{3}{16} \times \frac{9}{16}x^2 = \frac{27}{256}x^2\)

Thus, the expansion up to \(x^2\) is:

\(\frac{1}{2} - \frac{3}{16}x + \frac{27}{256}x^2\)