$1 \leftarrow \quad$ Solve the equation $48 x^{3}+40 x^{2}-x-3=0$ given that $-\frac{3}{4}$ is a zero of $f(x)=48 x^{3}+40 x^{2}-x-3$ The solution set is (Use a comma to separate answers as needed.)

Step 1 ：The question tells us that \(-\frac{3}{4}\) is a root of the equation. This means that if we substitute \(-\frac{3}{4}\) into the equation, it should equal to zero. We can use this information to factorize the equation and find the other roots.

Step 2 ：The factored form of the equation is \((3x + 1)(4x - 1)(4x + 3) = 0\). This gives us three roots of the equation: \(-\frac{1}{3}\), \(\frac{1}{4}\), and \(-\frac{3}{4}\). We already know that \(-\frac{3}{4}\) is a root, so the other two roots are \(-\frac{1}{3}\) and \(\frac{1}{4}\).

Step 3 ：Final Answer: The solution set is \(\boxed{-\frac{3}{4}, -\frac{1}{3}, \frac{1}{4}}\).