$$(2x^2-3x+5)(x^2+4x+1) = 0$$
Answer
$$ \begin{matrix}x_1 = -0.26795 & x_2 = -3.73205 & x_3 = 0.75+1.39194i \\[1 em] x_4 = 0.75-1.39194i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (2x^2-3x+5)(x^2+4x+1) &= 0&& \text{simplify left side} \\[1 em]2x^4+5x^3-5x^2+17x+5 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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Polynomial Equations Solver