If both the roots of the quadratic equation

$x^2+2(a+2)x+9a-1=0$ are negative, then $a$ lies in the set

if $D \ge 0$, sum of roots $< 0$and product of roots$> 0$

Thus, $4(a+2{)}^2-4(9a-1)\ge 0$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-2(a+2)<0,9a-1>0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2-5a+4\ge 0,a>-2,a>1/9\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(a-1)(a-4)\ge 0,a>1/9\end{array}$

$\Rightarrow$  $a\le 1$ or $a\ge 4$ and $a>1/9$

$\Rightarrow a\in (1/9,1]\cup [4,\mathrm{\infty })$