Set of all the values of $x$ satisfying the inequality $\sqrt{x^2-7x+6}>x+2$ is

$x^2-7x+6\ge 0\Rightarrow x\in \left(-\infty ,1\right]\cup \left[6,\infty \right)$
Case I: $x<-2$
LHS is non negative, RHS is negative
$\Rightarrow$ the inequality is always true

$\Rightarrow x<-2.$
Case II: $x\ge -2$
$x^2-7x+6>x^2+4x+4$
$-11x>-2\Rightarrow x<\frac{2}{11}$
$\Rightarrow x\in \left[-2,\frac{2}{11}\right)$
Taking the union of the two cases, we get, 
$x\in \left(-\infty ,\frac{2}{11}\right)$