$$r^4+2r^3-5r^2-3r+\frac{1}{2} = 0$$
Answer
$$ \begin{matrix}r_1 = 0.13715 & r_2 = -0.64321 & r_3 = 1.74818 \\[1 em] r_4 = -3.24212 & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} r^4+2r^3-5r^2-3r+\frac{1}{2} &= 0&& \text{multiply ALL terms by } \color{blue}{ 2 }. \\[1 em]2r^4+2\cdot2r^3-2\cdot5r^2-2\cdot3r+2\cdot\frac{1}{2} &= 2\cdot0&& \text{cancel out the denominators} \\[1 em]2r^4+4r^3-10r^2-6r+1 &= 0&& \\[1 em] \end{aligned} $$
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