$(2x+1)(x-3)=-6$
$\to 2x^2-6x+x-3=-6$
$\to 2x^2-5x-3+6=0$
$\to 2x^2-5x+3=0$
$\to 2x^2-3x-2x+3=0$
$\to x(2x-3)-(2x-3)=0$
$\to (2x-3)(x-1)=0$
$\to \left[ \begin{array}{l}2x-3=0\\x-1=0\end{array} \right.$
$\to \left[ \begin{array}{l}x=\dfrac{3}{2}\\x=1\end{array} \right.$
Vậy $\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=1\end{array} \right.$