$$x^4-15x^3+23x^2+44x = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -1.08685 & x_3 = 3.12278 \\[1 em] x_4 = 12.96406 & \\[1 em] \end{matrix} $$
Explanation
In order to solve $ \color{blue}{ x^{4}-15x^{3}+23x^{2}+44x = 0 } $, first we need to factor our $ x $.
$$ x^{4}-15x^{3}+23x^{2}+44x = x \left( x^{3}-15x^{2}+23x+44 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ x^{3}-15x^{2}+23x+44 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
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