$$(\frac{3}{2}x^2x-1)(2x^2-5x) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 5 }{ 2 } & x_3 = 0.87358 \\[1 em] x_4 = -0.43679+0.75654i & x_5 = -0.43679-0.75654i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (\frac{3}{2}x^2x-1)(2x^2-5x) &= 0&& \text{simplify left side} \\[1 em](\frac{3x^2}{2}x-1)(2x^2-5x) &= 0&& \\[1 em](\frac{3x^3}{2}-1)(2x^2-5x) &= 0&& \\[1 em]\frac{3x^3-2}{2}(2x^2-5x) &= 0&& \\[1 em]\frac{6x^5-15x^4-4x^2+10x}{2} &= 0&& \text{multiply ALL terms by } \color{blue}{ 2 }. \\[1 em]2 \cdot \frac{6x^5-15x^4-4x^2+10x}{2} &= 2\cdot0&& \text{cancel out the denominators} \\[1 em]6x^5-15x^4-4x^2+10x &= 0&& \\[1 em] \end{aligned} $$
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