Question
$$6x-\frac{4}{7}(21x-35x^2)+2 = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 3 }{ 20 }+\dfrac{\sqrt{ 31 }}{ 20 }i & x_2 = \dfrac{ 3 }{ 20 }-\dfrac{\sqrt{ 31 }}{ 20 }i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 6x-\frac{4}{7}(21x-35x^2)+2 &= 0&& \text{multiply ALL terms by } \color{blue}{ 7 }. \\[1 em]7\cdot6x-7 \cdot \frac{4}{7}(21x-35x^2)+7\cdot2 &= 7\cdot0&& \text{cancel out the denominators} \\[1 em]42x-(-140x^2+84x)+14 &= 0&& \text{simplify left side} \\[1 em]42x+140x^2-84x+14 &= 0&& \\[1 em]140x^2-42x+14 &= 0&& \\[1 em] \end{aligned} $$
$ 140x^{2}-42x+14 = 0 $ is a quadratic equation.
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