$$(5x^3+2x-3)(x^2-3x+6) = 0$$
Answer
$$ \begin{matrix}x_1 = 0.68753 & x_2 = -0.34376+0.86863i & x_3 = -0.34376-0.86863i \\[1 em] x_4 = 1.5+1.93649i & x_5 = 1.5-1.93649i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (5x^3+2x-3)(x^2-3x+6) &= 0&& \text{simplify left side} \\[1 em]5x^5-15x^4+32x^3-9x^2+21x-18 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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