$$10y^2-29 \cdot \frac{y}{2}+3 = 0$$
Answer
$$ \begin{matrix}y_1 = \dfrac{ 1 }{ 4 } & y_2 = \dfrac{ 6 }{ 5 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 10y^2-29 \cdot \frac{y}{2}+3 &= 0&& \text{multiply ALL terms by } \color{blue}{ 2 }. \\[1 em]2\cdot10y^2-2\cdot29 \cdot \frac{y}{2}+2\cdot3 &= 2\cdot0&& \text{cancel out the denominators} \\[1 em]20y^2-29y+6 &= 0&& \\[1 em] \end{aligned} $$
$ 20x^{2}-29x+6 = 0 $ is a quadratic equation.
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