Question
$$4+(3-5x\cdot2)x \cdot \frac{6-8}{7} = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 3 }{ 20 }+\dfrac{\sqrt{ 551 }}{ 20 }i & x_2 = \dfrac{ 3 }{ 20 }-\dfrac{\sqrt{ 551 }}{ 20 }i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 4+(3-5x\cdot2)x \cdot \frac{6-8}{7} &= 0&& \text{multiply ALL terms by } \color{blue}{ 7 }. \\[1 em]7\cdot4+7(3-5x\cdot2)x \cdot \frac{6-8}{7} &= 7\cdot0&& \text{cancel out the denominators} \\[1 em]28+20x^2-6x &= 0&& \text{simplify left side} \\[1 em]20x^2-6x+28 &= 0&& \\[1 em] \end{aligned} $$
$ 20x^{2}-6x+28 = 0 $ is a quadratic equation.
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