$$\frac{x^5}{5}-\frac{6}{8}x^4+\frac{9}{12}x^3-\frac{11}{4} = 0$$
Answer
$$ \begin{matrix}x_1 = 2.60947 & x_2 = 1.36676+1.35602i & x_3 = 1.36676-1.35602i \\[1 em] x_4 = -0.7965+0.88718i & x_5 = -0.7965-0.88718i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{x^5}{5}-\frac{6}{8}x^4+\frac{9}{12}x^3-\frac{11}{4} &= 0&& \text{multiply ALL terms by } \color{blue}{ 120 }. \\[1 em]120 \cdot \frac{x^5}{5}-120\frac{6}{8}x^4+120\frac{9}{12}x^3-120\cdot\frac{11}{4} &= 120\cdot0&& \text{cancel out the denominators} \\[1 em]24x^5-90x^4+90x^3-330 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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