Question
$$(5-3t)\cdot(5+3t) = 0$$
Answer
$$ \begin{matrix}t_1 = \dfrac{ 5 }{ 3 } & t_2 = -\dfrac{ 5 }{ 3 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (5-3t)\cdot(5+3t) &= 0&& \text{simplify left side} \\[1 em]25+15t-15t-9t^2 &= 0&& \\[1 em]25+15t-15t-9t^2 &= 0&& \\[1 em]-9t^2+25 &= 0&& \\[1 em] \end{aligned} $$
$ -9x^{2}+25 = 0 $ is a quadratic equation.
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