$$\frac{-6k^4+16k^3-9k^2}{3}k^2 = 0$$
Answer
$$ \begin{matrix}k_1 = 0 & k_2 = \dfrac{ 4 }{ 3 }-\dfrac{\sqrt{ 10 }}{ 6 } & k_3 = \dfrac{ 4 }{ 3 }+\dfrac{\sqrt{ 10 }}{ 6 } \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{-6k^4+16k^3-9k^2}{3}k^2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 3 }. \\[1 em]3 \cdot \frac{-6k^4+16k^3-9k^2}{3}k^2 &= 3\cdot0&& \text{cancel out the denominators} \\[1 em]-6k^6+16k^5-9k^4 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -6x^{6}+16x^{5}-9x^{4} = 0 } $, first we need to factor our $ x^4 $.
$$ -6x^{6}+16x^{5}-9x^{4} = x^4 \left( -6x^{2}+16x-9 \right) $$$ x = 0 $ is a root of multiplicity $ 4 $.
The remaining roots can be found by solving equation $ -6x^{2}+16x-9 = 0$.
$ -6x^{2}+16x-9 = 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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