$$t^2+\frac{9}{50}t-\frac{81}{10000} = 0$$
Answer
$$ \begin{matrix}t_1 = -\dfrac{ 9 }{ 100 }-\dfrac{ 9 \sqrt{ 2}}{ 100 } & t_2 = -\dfrac{ 9 }{ 100 }+\dfrac{ 9 \sqrt{ 2}}{ 100 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} t^2+\frac{9}{50}t-\frac{81}{10000} &= 0&& \text{multiply ALL terms by } \color{blue}{ 10000 }. \\[1 em]10000t^2+10000 \cdot \frac{9}{50}t-10000\cdot\frac{81}{10000} &= 10000\cdot0&& \text{cancel out the denominators} \\[1 em]10000t^2+1800t-81 &= 0&& \\[1 em] \end{aligned} $$
$ 10000x^{2}+1800x-81 = 0 $ is a quadratic equation.
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