$$4r(2r+3)(r-1) = 0$$
Answer
$$ \begin{matrix}r_1 = 0 & r_2 = 1 & r_3 = -\dfrac{ 3 }{ 2 } \end{matrix} $$
Explanation
$$ \begin{aligned} 4r(2r+3)(r-1) &= 0&& \text{simplify left side} \\[1 em](8r^2+12r)(r-1) &= 0&& \\[1 em]8r^3-8r^2+12r^2-12r &= 0&& \\[1 em]8r^3+4r^2-12r &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 8x^{3}+4x^{2}-12x = 0 } $, first we need to factor our $ x $.
$$ 8x^{3}+4x^{2}-12x = x \left( 8x^{2}+4x-12 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 8x^{2}+4x-12 = 0$.
$ 8x^{2}+4x-12 = 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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