$$\frac{5x}{2x^2-5x-7} = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} \frac{5x}{2x^2-5x-7} &= 0&& \text{multiply ALL terms by } \color{blue}{ 2x^2-5x-7 }. \\[1 em](2x^2-5x-7)\frac{5x}{2x^2-5x-7} &= (2x^2-5x-7)\cdot0&& \text{cancel out the denominators} \\[1 em]5x &= 0&& \text{ divide both sides by $ 5 $ } \\[1 em]x &= \frac{0}{5}&& \\[1 em]x &= 0&& \\[1 em] \end{aligned} $$
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