$$\frac{5}{2}+\frac{5}{3}(2x-1)\cdot14\frac{x}{3} = 0$$
Answer
$$ \begin{matrix}x_1 = \dfrac{ 1 }{ 4 }+\dfrac{\sqrt{ 77 }}{ 28 }i & x_2 = \dfrac{ 1 }{ 4 }-\dfrac{\sqrt{ 77 }}{ 28 }i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{5}{2}+\frac{5}{3}(2x-1)\cdot14\frac{x}{3} &= 0&& \text{multiply ALL terms by } \color{blue}{ 6 }. \\[1 em]6\cdot\frac{5}{2}+6\frac{5}{3}(2x-1)\cdot14\frac{x}{3} &= 6\cdot0&& \text{cancel out the denominators} \\[1 em]15+\frac{140}{3}(2x-1)x &= 0&& \text{multiply ALL terms by } \color{blue}{ 3 }. \\[1 em]3\cdot15+3 \cdot \frac{140}{3}(2x-1)x &= 3\cdot0&& \text{cancel out the denominators} \\[1 em]45+280x^2-140x &= 0&& \text{simplify left side} \\[1 em]280x^2-140x+45 &= 0&& \\[1 em] \end{aligned} $$
$ 280x^{2}-140x+45 = 0 $ is a quadratic equation.
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