Question
$$-2x(3x^2-5x+7) = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = \dfrac{ 5 }{ 6 }+\dfrac{\sqrt{ 59 }}{ 6 }i & x_3 = \dfrac{ 5 }{ 6 }-\dfrac{\sqrt{ 59 }}{ 6 }i \end{matrix} $$
Explanation
$$ \begin{aligned} -2x(3x^2-5x+7) &= 0&& \text{simplify left side} \\[1 em]-(6x^3-10x^2+14x) &= 0&& \\[1 em]-6x^3+10x^2-14x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -6x^{3}+10x^{2}-14x = 0 } $, first we need to factor our $ x $.
$$ -6x^{3}+10x^{2}-14x = x \left( -6x^{2}+10x-14 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ -6x^{2}+10x-14 = 0$.
$ -6x^{2}+10x-14 = 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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