$$-3v^3+4v^2-3v = 0$$
Answer
$$ \begin{matrix}v_1 = 0 & v_2 = \dfrac{ 2 }{ 3 }+\dfrac{\sqrt{ 5 }}{ 3 }i & v_3 = \dfrac{ 2 }{ 3 }-\dfrac{\sqrt{ 5 }}{ 3 }i \end{matrix} $$
Explanation
In order to solve $ \color{blue}{ -3x^{3}+4x^{2}-3x = 0 } $, first we need to factor our $ x $.
$$ -3x^{3}+4x^{2}-3x = x \left( -3x^{2}+4x-3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -3x^{2}+4x-3 = 0$.
$ -3x^{2}+4x-3 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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