$$6x(2x^3+4x)-2(x^2-8x^4) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{\sqrt{ 154 }}{ 14 } i & x_3 = - \dfrac{\sqrt{ 154 }}{ 14 } i \end{matrix} $$
Explanation
$$ \begin{aligned} 6x(2x^3+4x)-2(x^2-8x^4) &= 0&& \text{simplify left side} \\[1 em]12x^4+24x^2-(2x^2-16x^4) &= 0&& \\[1 em]12x^4+24x^2-2x^2+16x^4 &= 0&& \\[1 em]28x^4+22x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 28x^{4}+22x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 28x^{4}+22x^{2} = x^2 \left( 28x^{2}+22 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 28x^{2}+22 = 0$.
$ 28x^{2}+22 = 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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