$$\frac{9x^3-2x^2+6x-4}{5x^5-4x^4+3} = 0$$
Answer
$$ \begin{matrix}x_1 = 0.53364 & x_2 = -0.15571+0.89923i & x_3 = -0.15571-0.89923i \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{9x^3-2x^2+6x-4}{5x^5-4x^4+3} &= 0&& \text{multiply ALL terms by } \color{blue}{ 5x^5-4x^4+3 }. \\[1 em](5x^5-4x^4+3)\frac{9x^3-2x^2+6x-4}{5x^5-4x^4+3} &= (5x^5-4x^4+3)\cdot0&& \text{cancel out the denominators} \\[1 em]9x^3-2x^2+6x-4 &= 0&& \\[1 em] \end{aligned} $$
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