$$8a^8-20a^6+2 \cdot \frac{a^4}{4}a^3 = 0$$
Answer
$$ \begin{matrix}a_1 = 0 & a_2 = 0.48784 & a_3 = 1.57912 \\[1 em] a_4 = -1.58313 & a_5 = 0.14141+0.44968i & a_6 = 0.14141-0.44968i \\[1 em] a_7 = -0.38333+0.28928i & a_8 = -0.38333-0.28928i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 8a^8-20a^6+2 \cdot \frac{a^4}{4}a^3 &= 0&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4\cdot8a^8-4\cdot20a^6+42 \cdot \frac{a^4}{4}a^3 &= 4\cdot0&& \text{cancel out the denominators} \\[1 em]32a^8-80a^6+2a &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 32x^{8}-80x^{6}+2x = 0 } $, first we need to factor our $ x $.
$$ 32x^{8}-80x^{6}+2x = x \left( 32x^{7}-80x^{5}+2 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 32x^{7}-80x^{5}+2 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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