[tex]x^2+10x+24=(x+6)(x+4)\le0[/tex]
Notice that equality holds when either [tex]x=-6[/tex] or [tex]x=-4[/tex], so these should be included in the solution set.
Now, since the coefficient of the leading term is positive, we know the parabola opens upward, which means we know that [tex]x^2+10x+24>0[/tex] when [tex]x<-6[/tex] and [tex]x>-4[/tex], and that [tex]x^2+10x+24<0[/tex] when [tex]-6<x<-4[/tex].
So the solution set is the interval [tex]-6\le x\le-4[/tex].
