Question
$$5(t-1)^2+4(t-1) = 0$$
Answer
$$ \begin{matrix}t_1 = 1 & t_2 = \dfrac{ 1 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5(t-1)^2+4(t-1) &= 0&& \text{simplify left side} \\[1 em]5(1t^2-2t+1)+4(t-1) &= 0&& \\[1 em]5t^2-10t+5+4t-4 &= 0&& \\[1 em]5t^2-6t+1 &= 0&& \\[1 em] \end{aligned} $$
$ 5x^{2}-6x+1 = 0 $ is a quadratic equation.
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