[tex]\displaystyle\bf\\x^{2} +(a+5)x+1=0\\\\D=(a+5)^{2} -4\cdot 1=a^{2} +10a+25-4=a^{2} +10a+21\\\\D > 0 \ \ \Rightarrow \ \ a^{2} +10a+21 > 0\\\\(a+3)\cdot(a+7) > 0\\\\\\+ + + + + (-7) - - - - - (-3) + + + + + \\\\\\Otvet \ : \ pri \ \ a\in(-\infty \ ; \ -7)\cup(-3 \ ; \ +\infty)[/tex]