1. Consider the quadratic $f(x)=x^2-2kx+(k^2+k-5).$ We want both roots to be less than $5$.

2. First, ensure the quadratic has real roots. Its discriminant is $\Delta = (-2k)^2-4(k^2+k-5)=4k^2-4k^2-4k+20=20-4k=4(5-k).$ For real roots, $\Delta \ge 0 \implies 5-k\ge 0 \implies k\le 5.$

3. Let the roots be $\alpha,\beta$. Since the coefficient of $x^2$ is positive, for both roots to be less than $5$, it is necessary and sufficient that:

• roots are real,
• the larger root is $<5$.

A standard way is to shift the variable by $5$. Put $x=y+5.$ Then \begin{align*} f(y+5) &= (y+5)^2-2k(y+5)+k^2+k-5 \ &= y^2+(10-2k)y+(25-10k+k^2+k-5) \ &= y^2+(10-2k)y+(k^2-9k+20). \end{align*} So the new equation is $y^2+(10-2k)y+(k^2-9k+20)=0.$ For both original roots $x$ to be less than $5$, both new roots $y$ must be negative.

4. For a quadratic $y^2+py+q=0$ with real roots, both roots are negative iff

• sum of roots $=-p<0$,
• product of roots $=q>0$.

Here,

• sum of roots in $y$ is $-(10-2k)=2k-10,$ so for both negative: $2k-10<0 \implies k<5.$
• product of roots in $y$ is $k^2-9k+20=(k-4)(k-5),$ so $(k-4)(k-5)>0 \implies k<4 \text{ or } k>5.$

Combining with $k<5$, we get $k<4.$ Also this already satisfies the real-root condition $k\le 5$.

5. Hence, $k\in(-\infty,4).$

6. Checking options:

• A: $(5,6]$ — not possible
• B: $(6,\infty)$ — not possible
• C: $(-\infty,4)$ — correct
• D: $[4,5]$ — not possible

Therefore the correct option is C.