2.$x^{2}$ -5.x-3=0
⇔2$x^{2}$ +x-6x-3=0
⇔2x(x-3)+(x-3)=0
⇔(x-3)(2x+1)=0
⇒\(\left[ \begin{array}{l}x-3=0\\2x+1=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=3\\2x=-1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=3\\x=-1:2\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=3\\x=\frac{-1}{2}\end{array} \right.\)
Vậy x=3 hoặc x=$\frac{-1}{2}$