Answer
$\left( -\infty ,-1 \right)\cup \left( 2,\infty \right)\text{ or }\left\{ x|x2 \right\}$.
Work Step by Step
$f\left( x \right)={{x}^{2}}-x-2>0$
Now to find the interval, set the inequality to 0 and solve the equation.
$\begin{align}
& {{x}^{2}}-x-2=0 \\
& {{x}^{2}}-2x+x-2=0 \\
& x\left( x-2 \right)+1\left( x-2 \right)=0 \\
& \left( x-\left. 2 \right) \right.\left( x+\left. 1 \right) \right.=0
\end{align}$
The solution is,
$\begin{align}
& x-2=0 \\
& x=2
\end{align}$
Or,
$\begin{align}
& x+1=0 \\
& x=-1
\end{align}$
Now find out the sign of the equation ${{x}^{2}}-x-2$.
Take a test value from each interval and plug that into the equation:
For interval A,
Test for $x=-2$.
$\begin{align}
& {{\left( -2 \right)}^{2}}-\left( -2 \right)-2=4+4-2 \\
& =6>0
\end{align}$
So, in interval A, the equation ${{x}^{2}}-x-2>0$.
For interval B,
Test for $x=0$.
$\begin{align}
& {{\left( 0 \right)}^{2}}-\left( 0 \right)-2=-2 \\
& =-2\not{>}0
\end{align}$
So, in interval B, the equation ${{x}^{2}}-x-2\not{>}0$.
For interval C,
Test for $x=3$.
$\begin{align}
& {{\left( 3 \right)}^{2}}-\left( 3 \right)-2=9-3-2 \\
& =4>0
\end{align}$
So, in interval C, the equation ${{x}^{2}}-x-2$ is greater than 0.
Therefore, the solution of the inequality, ${{x}^{2}}-x-2>0$ is $\left( -\infty ,-1 \right)\cup \left( 2,\infty \right)\text{ or }\left\{ x|x2 \right\}$.
