a) -3x² + 15 = 0
↔ -3x² = -15
↔ x² = 5
↔ x = ±$\sqrt[]{5}$
Vậy S ={±$\sqrt[]{5}$}
b) x² - 3x - 10 = 0
↔ x² + 2x - 5x - 10 = 0
↔ x(x + 2) - 5(x + 2) = 0
↔ (x + 2)(x - 5) = 0
↔ \(\left[ \begin{array}{l}x+2 = 0\\x-5 = 0\end{array} \right.\)
↔ \(\left[ \begin{array}{l}x = -2\\x = 5\end{array} \right.\)
Vậy S ={-2; 5}