$$(4a^2-5a+7)(6a^2+3a-2) = 0$$
Answer
$$ \begin{matrix}a_1 = 0.37915 & a_2 = -0.87915 & a_3 = 0.625+1.16592i \\[1 em] a_4 = 0.625-1.16592i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (4a^2-5a+7)(6a^2+3a-2) &= 0&& \text{simplify left side} \\[1 em]24a^4-18a^3+19a^2+31a-14 &= 0&& \\[1 em] \end{aligned} $$
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