<s>[INST] What is the smallest value of $x$ such that the matrix $A$ is not invertible? 
$$A = \begin{pmatrix}
-1 & -3 & x \\
-2 & -7 & -1 \\
x & -6 & 1 
\end{pmatrix}$$ 

So far I tried taking the determinant by expanding down the last column but i still cant get the right answer.
x * det $ \begin{pmatrix}
-2 & -7 \\x & -6 
\end{pmatrix} $ + 1 * det  $ \begin{pmatrix}
-1 & -3 \\
x & -6
\end{pmatrix} $ +1*det $ \begin{pmatrix}
-1 & -3 \\
-2 & -7 
\end{pmatrix} $
I end  up with $7x^2+15x+7$$=$$0$ 
 [/INST] You're on the right way continue by finding the roots
$x_1=-\frac{15}{14} - \frac{\sqrt{29}}{14}$
$x_2=-\frac{15}{14} + \frac{\sqrt{29}}{14}$
And taking the $\min\{x_1,x_2\} = -\frac{15}{14} - \frac{\sqrt{29}}{14}$
That's it that is your answer
</s>