$$(3x-1)(2x^3+4x^2-5) = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 3 } & x_2 = 0.92457 & x_3 = -1.46228+0.75212i \\[1 em] x_4 = -1.46228-0.75212i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (3x-1)(2x^3+4x^2-5) &= 0&& \text{simplify left side} \\[1 em]6x^4+12x^3-15x-2x^3-4x^2+5 &= 0&& \\[1 em]6x^4+10x^3-4x^2-15x+5 &= 0&& \\[1 em] \end{aligned} $$
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