$$x(x-\frac{1}{2})(x+3) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 1 }{ 2 } & x_3 = -3 \end{matrix} $$
Explanation
$$ \begin{aligned} x(x-\frac{1}{2})(x+3) &= 0&& \text{simplify left side} \\[1 em]x \cdot \frac{2x-1}{2}(x+3) &= 0&& \\[1 em]\frac{2x^2-x}{2}(x+3) &= 0&& \\[1 em]\frac{2x^3+5x^2-3x}{2} &= 0&& \text{multiply ALL terms by } \color{blue}{ 2 }. \\[1 em]2 \cdot \frac{2x^3+5x^2-3x}{2} &= 2\cdot0&& \text{cancel out the denominators} \\[1 em]2x^3+5x^2-3x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 2x^{3}+5x^{2}-3x = 0 } $, first we need to factor our $ x $.
$$ 2x^{3}+5x^{2}-3x = x \left( 2x^{2}+5x-3 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 2x^{2}+5x-3 = 0$.
$ 2x^{2}+5x-3 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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