$$-4j^2(3j^2+4j-3) = 0$$
Answer
$$ \begin{matrix}j_1 = 0 & j_2 = -\dfrac{ 2 }{ 3 }-\dfrac{\sqrt{ 13 }}{ 3 } & j_3 = -\dfrac{ 2 }{ 3 }+\dfrac{\sqrt{ 13 }}{ 3 } \end{matrix} $$
Explanation
$$ \begin{aligned} -4j^2(3j^2+4j-3) &= 0&& \text{simplify left side} \\[1 em]-(12j^4+16j^3-12j^2) &= 0&& \\[1 em]-12j^4-16j^3+12j^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -12x^{4}-16x^{3}+12x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ -12x^{4}-16x^{3}+12x^{2} = x^2 \left( -12x^{2}-16x+12 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ -12x^{2}-16x+12 = 0$.
$ -12x^{2}-16x+12 = 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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