$$7a^5+a-3a^2+3 = 0$$
Answer
$$ \begin{matrix}a_1 = -0.67315 & a_2 = 0.7307+0.42739i & a_3 = 0.7307-0.42739i \\[1 em] a_4 = -0.39412+0.85624i & a_5 = -0.39412-0.85624i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 7a^5+a-3a^2+3 &= 0&& \text{simplify left side} \\[1 em]7a^5-3a^2+a+3 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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