$$\frac{6x^3+4x^2-8x+3}{x+1} = 0$$
Answer
$$ \begin{matrix}x_1 = -1.65491 & x_2 = 0.49412+0.24077i & x_3 = 0.49412-0.24077i \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{6x^3+4x^2-8x+3}{x+1} &= 0&& \text{multiply ALL terms by } \color{blue}{ x+1 }. \\[1 em](x+1)\frac{6x^3+4x^2-8x+3}{x+1} &= (x+1)\cdot0&& \text{cancel out the denominators} \\[1 em]6x^3+4x^2-8x+3 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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Equations Solver