$$-4j\cdot2(3j\cdot2+4j-3) = 0$$
Answer
$$ \begin{matrix}j_1 = 0 & j_2 = \dfrac{ 3 }{ 10 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} -4j\cdot2(3j\cdot2+4j-3) &= 0&& \text{simplify left side} \\[1 em]-8j(6j+4j-3) &= 0&& \\[1 em]-8j(10j-3) &= 0&& \\[1 em]-(80j^2-24j) &= 0&& \\[1 em]-80j^2+24j &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -80x^{2}+24x = 0 } $, first we need to factor our $ x $.
$$ -80x^{2}+24x = x \left( -80x+24 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -80x+24 = 0$.
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