$$\frac{-24s^5+42s^4+48s^3}{6s^2} = 0$$
Answer
$$ \begin{matrix}s_1 = 0 & s_2 = \dfrac{ 7 }{ 8 }-\dfrac{\sqrt{ 177 }}{ 8 } & s_3 = \dfrac{ 7 }{ 8 }+\dfrac{\sqrt{ 177 }}{ 8 } \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{-24s^5+42s^4+48s^3}{6s^2} &= 0&& \text{multiply ALL terms by } \color{blue}{ 6s^2 }. \\[1 em]6s^2\frac{-24s^5+42s^4+48s^3}{6s^2} &= 6s^2\cdot0&& \text{cancel out the denominators} \\[1 em]-24s^9+42s^8+48s^7 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -24x^{9}+42x^{8}+48x^{7} = 0 } $, first we need to factor our $ x^7 $.
$$ -24x^{9}+42x^{8}+48x^{7} = x^7 \left( -24x^{2}+42x+48 \right) $$$ x = 0 $ is a root of multiplicity $ 7 $.
The remaining roots can be found by solving equation $ -24x^{2}+42x+48 = 0$.
$ -24x^{2}+42x+48 = 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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