Question
$$\frac{t^3-4t^2}{t^2+2t-15} = 0$$
Answer
$$ \begin{matrix}t_1 = 0 & t_2 = 4 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{t^3-4t^2}{t^2+2t-15} &= 0&& \text{multiply ALL terms by } \color{blue}{ t^2+2t-15 }. \\[1 em](t^2+2t-15)\frac{t^3-4t^2}{t^2+2t-15} &= (t^2+2t-15)\cdot0&& \text{cancel out the denominators} \\[1 em]t^3-4t^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ x^{3}-4x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ x^{3}-4x^{2} = x^2 \left( x-4 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The second root can be found by solving equation $ x-4 = 0$.
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