$$(3x^2-4x-5)(2x^6-5) = 0$$
Answer
$$ \begin{matrix}x_1 = -0.7863 & x_2 = 1.16499 & x_3 = -1.16499 \\[1 em] x_4 = 2.11963 & x_5 = 0.5825+1.00891i & x_6 = 0.5825-1.00891i \\[1 em] x_7 = -0.5825+1.00891i & x_8 = -0.5825-1.00891i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (3x^2-4x-5)(2x^6-5) &= 0&& \text{simplify left side} \\[1 em]6x^8-15x^2-8x^7+20x-10x^6+25 &= 0&& \\[1 em]6x^8-8x^7-10x^6-15x^2+20x+25 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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