Some doubts with the sign of a derivative Good evening to everyone. The derivative is defined in the following order:
$$ \frac{d}{dx} f(x)=\frac{{-x^2-x+11}}{\left(x+3\right)^2}e^{2-x}\:$$ for $ x < -3 $ $$\:\frac{d}{dx}f(x)=\frac{{x^2+x-11}}{\left(x+3\right)^2}e^{2-x} $$ for $ -3< x< 2 $ and $$ \frac{d}{dx}f(x) = \frac{{x^2+x+1}}{\left(x+3\right)^2}e^{x-2} $$ for $x>2$. If you'll do the sign of the first one it'll be $$ {x^2+x-11} < 0 $$ therefore only $$ x_1 = \frac{-1+\sqrt{45}}{2}  $$ verifies the conditions, for the second one it is $$ {x^2+x-11} > 0 $$ therefore $x_1 = \frac{-1+\sqrt{45}}{2}$ and $ x_1 = \frac{-1-\sqrt{45}}{2}$ , the last one $$ {x^2+x+1} > 0  $$ has the solutions $ x_1= \frac{-1-i\sqrt{3}}{2}$ and $ x_2 = \frac{-1+i\sqrt{3}}{2} $(here I think that I did a mistake). Therefore our function should increase on the interval $ (-\infty, \frac{-1-\sqrt{45}}{2}) $ then decrease on the interval $ (\frac{-1-\sqrt{45}}{2},\frac{-1+\sqrt{45}}{2}) $ then increase again on the interval $(\frac{-1+\sqrt{45}}{2}, \frac{-1-i\sqrt{3}}{2})$ and decrease again on $(\frac{-1+i\sqrt{3}}{2},\infty)$. But Instead the solutions of the third equation are $x_1 = \frac{-1-\sqrt{5}}{2} $ and $ x_2 = \frac{-1+\sqrt{5}}{2}$. And it is decreasing on $ (-\infty, \frac{-1-\sqrt{45}}{2}) $, then increasing on $ (\frac{-1-\sqrt{45}}{2}, -3) $ then decreasing on $(-3,2)$ then increasing again on $ (2,\infty)$. I don't get where I'm wrong.. neither with the results of quadratic equation nor with the intervals of monotony(sign) of the derivative.

We have a root for the first derivative at around $x\approx-3.854$ the other is negligible as the derivative is not defined with the particular function at that point. 
With $x<-\frac{1+3\sqrt{5}}{2}, f'(x)<0 $. Thus, $f(x)$ is decreasing for  $x\in(-\infty,-\frac{1+3\sqrt{5}}{2})$ and increasing for $x\in( -\frac{1+3\sqrt{5}}{2},-3)$. (I substituted a random value for $x<-\frac{1+3\sqrt{5}}{2}$ and found that $f'(x)<0$)
For $-3<x<2$, the derivative defined as the same roots so now the root around $x\approx2.854$ is relevant and the other is negligible.
With $-3<x<2, f'(x)$ is negative. Thus, $f(x)$ is decreasing from $x\in(-3,2)$.
For $x>2$ the derivative defined does not have any real roots. With a random check, we find that at any point $x>2, f'(x)>0$. Thus, $f(x)$ is always increasing on $x\in(2,\infty)$.
