$$(3v+1)(2v-5) = 0$$
Answer
$$ \begin{matrix}v_1 = -\dfrac{ 1 }{ 3 } & v_2 = \dfrac{ 5 }{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (3v+1)(2v-5) &= 0&& \text{simplify left side} \\[1 em]6v^2-15v+2v-5 &= 0&& \\[1 em]6v^2-13v-5 &= 0&& \\[1 em] \end{aligned} $$
$ 6x^{2}-13x-5 = 0 $ is a quadratic equation.
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