-x² + 3x - 2 = 0
⇔ -x² + 2x + x - 2 = 0
⇔ (-x²+2x) + (x-2) = 0
⇔ -x(x - 2) + (x-2) = 0
⇔ (x-2)(1-x) = 0
⇔ \(\left[ \begin{array}{l}x-2=0\\1-x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)
Vậy pt đã cho có nghiệm là: \(\left[ \begin{array}{l}x=2\\x=1\end{array} \right.\)