If each of the roots of the equation $x^2-(a+1)x+a^2+a-8=0$ is greater than 1, then the range of '$a$’ is

$f\left(x\right)=x^2-(a+1)x+a^2+a-8=0$ has real roots $(i.e.,\Delta \ge 0)$ And $f\left(1\right)>0and\left(\frac{a+1}{2}\right)>1$

$\frac{-11}{3}\le a\le 3 and (a<-2\sqrt{2}ora>2\sqrt{2})\&a>1$

$\Rightarrow 2\sqrt{2}<a\le 3$