To solve the equation \( 4(3x + 4)^2 - 3(3x + 4) - 1 = 0 \), we can use substitution to simplify our calculations.

Let \( y = 3x + 4 \). Then, our equation becomes:

\[
4y^2 - 3y - 1 = 0
\]

Now, we can apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -3 \), and \( c = -1 \).

First, calculate the discriminant:

\[
b^2 - 4ac = (-3)^2 - 4(4)(-1) = 9 + 16 = 25
\]

Now substitute into the quadratic formula:

\[
y = \frac{-(-3) \pm \sqrt{25}}{2 \cdot 4} = \frac{3 \pm 5}{8}
\]

This gives us two solutions for \( y \):

1. For the plus:

\[
y = \frac{3 + 5}{8} = \frac{8}{8} = 1
\]

2. For the minus:

\[
y = \frac{3 - 5}{8} = \frac{-2}{8} = -\frac{1}{4}
\]

Now substitute back for \( y \):

1. For \( y = 1 \):

\[
3x + 4 = 1 \implies 3x = 1 - 4 \implies 3x = -3 \implies x = -1
\]

2. For \( y = -\frac{1}{4} \):

\[
3x + 4 = -\frac{1}{4} \implies 3x = -\frac{1}{4} - 4 \implies 3x = -\frac{1}{4} - \frac{16}{4} \implies 3x = -\frac{17}{4}
\]

Now solving for \( x \):

\[
x = -\frac{17}{12}
\]

Thus, the solutions to the original equation \( 4(3x + 4)^2 - 3(3x + 4) - 1 = 0 \) are:

\[
x = -1 \quad \text{and} \quad x = -\frac{17}{12}
\]