If $x =-9$ is a root of $\left|\begin{array}{l}x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}7\\ 2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ 7\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}6\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x\end{array}\right|=0$ then the other two roots are

Applying , $R_1\to R_1+R_2+R_3$ we get 

$\mathrm{\Delta }=\left|\begin{array}{ccc}x+9& x+9& x+9\\ 2& x& 2\\ \overline{7}& 6& x\end{array}\right|$

Taking x + 9 common from R1 and applying $C_2 \to C_{2 }– C_1, C_3 \to {C_3}_{ }– C_1$ we get

$\mathrm{\Delta }=(x+9)\left|\begin{array}{ccc}1& 0& 0\\ 2& x-2& 0\\ 7& -1& x-7\end{array}\right|$

$= (x + 9) (x – 2) (x – 7)$

Thus , $∆ = 0\Rightarrow x =– 9, 2, 7$

$\therefore$ the other two roots are 2 and 7.