Linear Algebra: solving for x

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 $x=-1$

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The answer is correct but there is a sign error in the 2nd equation, which is corrected in the 4th equation. – Bill Dubuque Jun 9 '11 at 15:24

There should be. A second error saves the day later on, when they transform $x^2-2x+3=0$ (false) into $(x-3)(x+1)=0$ (true again).
This is a mistake. To calculate the determinant of a $2\times 2$ matrix you use $ad-bc$ which gives you $(x-2)x-(-1)(-3)=x^2-2x-3=(x+1)(x-3)=0$. Hence $x=3$ or $x=-1$.