$$-3a^2(-3a^3-a-6) = 0$$
Answer
$$ \begin{matrix}a_1 = 0 & a_2 = -1.17189 & a_3 = 0.58594+1.16761i \\[1 em] a_4 = 0.58594-1.16761i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} -3a^2(-3a^3-a-6) &= 0&& \text{simplify left side} \\[1 em]-(-9a^5-3a^3-18a^2) &= 0&& \\[1 em]9a^5+3a^3+18a^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 9x^{5}+3x^{3}+18x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 9x^{5}+3x^{3}+18x^{2} = x^2 \left( 9x^{3}+3x+18 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 9x^{3}+3x+18 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
This page was created using
Equations Solver