I was wondering why in this exercise there isn't a minus $(x^2-2x-3)$. I thought that for this type of exercise you must use $(ad-bc)$.
Problem 50 (p 131) Solve for $x$.
$\begin{vmatrix} x-2& -1\\ -3 & x \end{vmatrix}=0$
$(x-2)\cdot x + (-3)\cdot(-1)=0$
$x^2-2x+3=0$
$x-3=0$ or $x+1=0$
$x=3$ or $x=-1$