$$(x^2+1)(x^2-3x+5) = 0$$
Answer
$$ \begin{matrix}x_1 = 1i & x_2 = -i & x_3 = 1.5+1.65831i \\[1 em] x_4 = 1.5-1.65831i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x^2+1)(x^2-3x+5) &= 0&& \text{simplify left side} \\[1 em]x^4-3x^3+5x^2+x^2-3x+5 &= 0&& \\[1 em]x^4-3x^3+6x^2-3x+5 &= 0&& \\[1 em] \end{aligned} $$
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