$$5x^2+3x^3-x^4+\frac{1}{7} = 0$$
Answer
$$ \begin{matrix}x_1 = -1.21062 & x_2 = 4.19409 & x_3 = 0.00827+0.16753i \\[1 em] x_4 = 0.00827-0.16753i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5x^2+3x^3-x^4+\frac{1}{7} &= 0&& \text{multiply ALL terms by } \color{blue}{ 7 }. \\[1 em]7\cdot5x^2+7\cdot3x^3-7x^4+7\cdot\frac{1}{7} &= 7\cdot0&& \text{cancel out the denominators} \\[1 em]35x^2+21x^3-7x^4+1 &= 0&& \text{simplify left side} \\[1 em]-7x^4+21x^3+35x^2+1 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
This page was created using
Polynomial Equations Solver