↔ $\frac{-(x^2-4x+3)}{2}$ + $\frac{4(x-1)}{2}$ = 0
↔ -($x^{2}$ -4x + 3) + 4(x - 1) = 0
↔ -$x^{2}$ + 4x - 3 + 4(x - 1) = 0
↔ $x^{2}$ - 4x + 3 - 4(x - 1) = 0
↔ $x^{2}$ - x - 3x + 3 - 4(x - 1) = 0
↔ x( x - 1) - 3(x - 1) - 4(x - 1) = 0
↔ (x - 1)(x - 3) - 4(x-1) = 0
↔ (x - 1)(x - 3 -4) = 0
↔ (x - 1)(x - 7) = 0
↔ \(\left[ \begin{array}{l}x-1=0\\x-7=0\end{array} \right.\)
↔ \(\left[ \begin{array}{l}x=1\\x=7\end{array} \right.\)
Vậy S ={1; 7}
