Official solution
Case 1 :
x \[ \le \] -4
(-x - 3)(-x - 4) = 6
\[ \Rightarrow \] (x + 3)(x + 4) = 6
\[ \Rightarrow \] x2 + 7x + 6 = 0
\[ \Rightarrow \] x = -1 or -6
but x \[ \le \] -4
x = -6
Case 2 :
x \[\in\] (-4, 0)
(-x - 3)(x + 4) = 6
\[ \Rightarrow \] -x2 - 7x - 12 - 6 = 0
\[ \Rightarrow \] x2 + 7x + 18 = 0
D < 0 No solution
Case 3 :
x \[ \ge \] 0
(x - 3)(x + 4) = 6
\[ \Rightarrow \] x2 + x - 12 - 6 = 0
\[ \Rightarrow \] x2 + x - 18 = 0
x = \[{{ - 1 \pm \sqrt {1 + 72} } \over 2}\]
\[ \therefore \] x = \[{{\sqrt {73} - 1} \over 2}\] only