$$\frac{12x^3+20x^2-22x-15}{6}x+4 = 0$$
Answer
$$ \begin{matrix}x_1 = -1.21565 & x_2 = -2.02668 & x_3 = 0.78783+0.43714i \\[1 em] x_4 = 0.78783-0.43714i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{12x^3+20x^2-22x-15}{6}x+4 &= 0&& \text{multiply ALL terms by } \color{blue}{ 6 }. \\[1 em]6 \cdot \frac{12x^3+20x^2-22x-15}{6}x+6\cdot4 &= 6\cdot0&& \text{cancel out the denominators} \\[1 em]12x^4+20x^3-22x^2-15x+24 &= 0&& \\[1 em] \end{aligned} $$
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