$$(2x+3)(4x^5+2x^4-4x^2+4x+8) = 0$$
Answer
$$ \begin{matrix}x_1 = -\dfrac{ 3 }{ 2 } & x_2 = -0.90592 & x_3 = 0.91968+0.65981i \\[1 em] x_4 = 0.91968-0.65981i & x_5 = -0.71672+1.09978i & x_6 = -0.71672-1.09978i \end{matrix} $$
Explanation
$$ \begin{aligned} (2x+3)(4x^5+2x^4-4x^2+4x+8) &= 0&& \text{simplify left side} \\[1 em]8x^6+16x^5+6x^4-8x^3-4x^2+28x+24 &= 0&& \\[1 em] \end{aligned} $$
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