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($\dfrac{1}{2}$ `- x) (x +` $\dfrac{3}{5}$ `) = 0`
⇔ \(\left[ \begin{array}{l}\dfrac{1}{2} - x=0\\x + \dfrac{3}{5}=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l} x=\dfrac{1}{2}\\x + \dfrac{3}{5}=- \dfrac{3}{5}\end{array} \right.\)
Vậy `x ∈ {` $\dfrac{1}{2}$ ; $\dfrac{-3}{5}$ }