Linear Algebra: different determinant answers I'm having a problem verifying my answer to this question:
Solve for x:
$$\left| \begin{array}{cc}
x+3 & 2 \\
1 & x+2 \end{array} \right| = 0$$
I get:
$(x+3)(x+2)-2=0$ 
$(x+3)(x+2)=2$
thus:

$x+3=2$ and    $x+2=2$

$x=-1$ and $x=0$
The book says that $x=-1$ and $x=-4$ is the correct answer.
I tried doing it a different way by expanding and got totally different answers:
$x^2+5x=-4$
$x(x+5)=-4$
$x=-4$ and $x=-9$
What is going on here?

The step in your reasoning
$$(x+3)(x+2)=2 $$
$$\Longrightarrow$$
$$x+3=2\text{ and }x+2=2$$
is incorrect - if $x+2=2$, then $x=0$, hence $x+3=0+3=3$, hence $(x+3)(x+2)=3\cdot 2=6$, not $2$. Also the step in your second approach
$$x^2+5x=-4$$
$$\Longrightarrow$$
$$x(x+5)=-2$$
is incorrect. $x^2+5x$ and $x(x+5)$ are the same thing, so it can't equal both $-4$ and $-2$.
Your second approach is on the right track though; you are correct in multiplying $(x+3)$ and $(x+2)$ together, but now bring everything over to one side, leaving $0$ on the other side:
$$x^2+5x+4=0$$
Now use the quadratic formula to get that the solutions are $x=-1$ and $x=-4$.
