if $s=\sqrt{x^2+6x+9} +\sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that... 
if $s=\sqrt{x^2+6x+9} + \sqrt{x^2+24x+144}$ and $-10 < x < -8$ there follows that $s=9$

After noticing that the first radical is $x+3$ and the second one is  $x+12$ all I get is $s=2x+15$, which btw. is suggested as an alternative answer in the exam.
So, how to go about this?

Note that $$x^2+6x+9=(x+3)^2$$ and $$x^2+24x+144=(x+12)^2$$, so you will get
$$s=|x+3|+|x+12|$$, now consider the inequality $$-10<x<-8$$
