To solve the equation \(3x^3 - x^2 + 4x = 0\), we can start by factoring out the greatest common factor (GCF) from each term.

The GCF of \(3x^3\), \(-x^2\), and \(4x\) is \(x\). So, we factor \(x\) out of the equation:

\[ x(3x^2 - x + 4) = 0 \]

This gives us two factors to consider:

1. \( x = 0 \)
2. \( 3x^2 - x + 4 = 0 \)

The first factor \( x = 0 \) is straightforward and gives us one solution:

\[ x = 0 \]

Next, we need to solve the quadratic equation \( 3x^2 - x + 4 = 0 \). We can use the quadratic formula, which is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For the quadratic equation \( 3x^2 - x + 4 = 0 \), the coefficients are:
- \( a = 3 \)
- \( b = -1 \)
- \( c = 4 \)

Plugging these values into the quadratic formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \]
\[ x = \frac{1 \pm \sqrt{1 - 48}}{6} \]
\[ x = \frac{1 \pm \sqrt{-47}}{6} \]

Since the discriminant (\(1 - 48 = -47\)) is negative, the quadratic equation has no real solutions. Instead, it has two complex solutions. We can express the solutions in terms of imaginary numbers:

\[ x = \frac{1 \pm \sqrt{47}i}{6} \]

So, the complete set of solutions to the equation \(3x^3 - x^2 + 4x = 0\) is:

\[ x = 0 \]
\[ x = \frac{1 + \sqrt{47}i}{6} \]
\[ x = \frac{1 - \sqrt{47}i}{6} \]