$$(t-1)(4t-2) = 0\cdot0$$
Answer
$$ \begin{matrix}t_1 = 1 & t_2 = \dfrac{ 1 }{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (t-1)(4t-2) &= 0\cdot0&& \text{simplify left and right hand side} \\[1 em]4t^2-2t-4t+2 &= 0&& \\[1 em]4t^2-6t+2 &= 0&& \\[1 em] \end{aligned} $$
$ 4x^{2}-6x+2 = 0 $ is a quadratic equation.
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