Question
$$6x^2+3x+\frac{4}{2}x^2-3x-2 = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 2 } & x_2 = -\dfrac{ 1 }{ 2 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 6x^2+3x+\frac{4}{2}x^2-3x-2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 2 }. \\[1 em]2\cdot6x^2+2\cdot3x+2 \cdot \frac{4}{2}x^2-2\cdot3x-2\cdot2 &= 2\cdot0&& \text{cancel out the denominators} \\[1 em]12x^2+6x+4x^2-6x-4 &= 0&& \text{simplify left side} \\[1 em]12x^2+6x+4x^2-6x-4 &= 0&& \\[1 em]16x^2-4 &= 0&& \\[1 em] \end{aligned} $$
$ 16x^{2}-4 = 0 $ is a quadratic equation.
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