$$(6x^8-4x^5-1)(3x^2-4) = 0$$
Answer
$$ \begin{matrix}x_1 = 0.95643 & x_2 = -0.69791 & x_3 = 1.1547 \\[1 em] x_4 = -1.1547 & x_5 = 0.54574+0.45879i & x_6 = 0.54574-0.45879i \\[1 em] x_7 = -0.12802+0.76263i & x_8 = -0.12802-0.76263i & x_9 = -0.54698+0.72266i \\[1 em] x_10 = -0.54698-0.72266i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (6x^8-4x^5-1)(3x^2-4) &= 0&& \text{simplify left side} \\[1 em]18x^{10}-24x^8-12x^7+16x^5-3x^2+4 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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