Question
$$\frac{5x^2-18x+24}{x^3-4x^2+4x} = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 9 }{ 5 }+\dfrac{\sqrt{ 39 }}{ 5 }i & x_2 = \dfrac{ 9 }{ 5 }-\dfrac{\sqrt{ 39 }}{ 5 }i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{5x^2-18x+24}{x^3-4x^2+4x} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^3-4x^2+4x }. \\[1 em](x^3-4x^2+4x)\frac{5x^2-18x+24}{x^3-4x^2+4x} &= (x^3-4x^2+4x)\cdot0&& \text{cancel out the denominators} \\[1 em]5x^2-18x+24 &= 0&& \\[1 em] \end{aligned} $$
$ 5x^{2}-18x+24 = 0 $ is a quadratic equation.
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