$$a^3+5a+11-(4a^3+6a^2-a+1) = 0$$
Answer
$$ \begin{matrix}a_1 = 1.34183 & a_2 = -1.11613 & a_3 = -2.2257 \end{matrix} $$
Explanation
$$ \begin{aligned} a^3+5a+11-(4a^3+6a^2-a+1) &= 0&& \text{simplify left side} \\[1 em]a^3+5a+11-4a^3-6a^2+a-1 &= 0&& \\[1 em]-3a^3-6a^2+6a+10 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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Polynomial Equations Solver