`( 3x + 2 ) . ( 5 - x^2 ) = 0`
`⇔` \(\left[ \begin{array}{l}3x + 2 = 0\\5 - x² = 0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}3x = 2\\x^2 = 5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = \frac{2}{3} \\x = +√5 ; x = -√5\end{array} \right.\)
Vậy , `x ∈ { 2/3 ; +√5 ; -√5 }`