Solve the folowing inequality over the real domain:
\( x^2-x-2 \le 0 \)
\( x_{1{,}2}=\frac{-1\pm\sqrt{1+8}}{2}=\frac{-1\pm 3}{2} \)
Since the quadratic term has a positive coefficient, \(x^2-x-2\) is positive when the absolute value of x is large.
The solution is \(x\in \langle-2{,}1\rangle\).
