$Ifbothrootsofthequadraticequation{x}^2-2kx+k^2+k-5=0arelessthan5,thenkliesinthe\mathrm{int}erval$

$\begin{array}{l}Giveequation{x}^2-2kx+(k^2+k-5)=0\\ \therefore x=\frac{2k\pm \sqrt{4k^2-4(k^2+k-5)}}{2\left(1\right)}\\ =\frac{2k\pm \sqrt{4k^2-4k^2-4k+20}}{2}\\ =\frac{2k\pm \sqrt{20-4k}}{2}\\ =\frac{2k\pm 2\sqrt{5-k}}{2}=k\pm \sqrt{5-k}\\ Givenbothrootsarelessthan5\\ \therefore k\pm \sqrt{5-k}<5\\ \Rightarrow \pm \sqrt{5-k}<5-k\\ \Rightarrow 5-k<{(5-k)}^2\\ \Rightarrow 1<(5-k)\\ \therefore k<4\Rightarrow k\in (-\infty ,4)\end{array}$