$$(t-1)(4t-2)\cdot0 = 0$$
Answer
The equation has an infinite number of solutions.
Explanation
$$ \begin{aligned} (t-1)(4t-2)\cdot0 &= 0&& \text{simplify left side} \\[1 em](4t^2-2t-4t+2)\cdot0 &= 0&& \\[1 em](4t^2-6t+2)\cdot0 &= 0&& \\[1 em]0t^2+0t+0 &= 0&& \\[1 em]0 &= 0&& \\[1 em] \end{aligned} $$
Since the statement $ \color{blue}{ 0 = 0 } $ is TRUE for any value of $ x $, we conclude that the equation has infinitely many solutions.
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