Cách giải:
$i,(x-4)(x+2)=2$
$\to x^2-4x+2x-8=2$
$\to x^2-2x-8=2$
$\to x^2-2x=10$
$\to x^2-2x+1=11$
$\to (x-1)^2=11$
$\to \left[ \begin{array}{l}x-1=\sqrt{11}\\x-1=-\sqrt{11}\end{array} \right.$
$\to \left[ \begin{array}{l}x=1+\sqrt{11}\\x=1-\sqrt{11}\end{array} \right.$
$b,4(x+1)-(3x+2)=14$
$\to 4x+4-3x-2=14$
$\to x+2=14$
$\to x=12$
Vậy $x=12$