Given quadratic equation

$x^2+4x-n=0$

We need the values of natural number $n \in [20,100]$ such that the equation has integral roots.

Condition for integral roots

For the quadratic $x^2+4x-n=0,$ if roots are integers, then the discriminant must be a perfect square.

Discriminant: $D=b^2-4ac=4^2-4(1)(-n)=16+4n=4(n+4)$

So, $D=4(n+4)$

For roots to be integers, $\sqrt{D}=\sqrt{4(n+4)}=2\sqrt{n+4}$ must be an integer. Hence $n+4$ must be a perfect square.

Let $n+4=k^2$ where $k$ is a positive integer.

Then $n=k^2-4$

Apply the range of $n$

Since $20 \le n \le 100,$ we get $20 \le k^2-4 \le 100$

Adding $4$ throughout: $24 \le k^2 \le 104$

Now list perfect squares in this interval: $25,36,49,64,81,100$

Thus possible values of $k^2$ are these 6 numbers.

Corresponding values of $n$

$n=25-4=21$ $n=36-4=32$ $n=49-4=45$ $n=64-4=60$ $n=81-4=77$ $n=100-4=96$

Total number of distinct values of $n$ is $6$

Check roots are indeed integers

Roots are $x=\frac{-4\pm\sqrt{16+4n}}{2}=\frac{-4\pm 2k}{2}=-2\pm k$

Since $k$ is an integer, both roots are integers. So all above values are valid.

Conclusion

The required number of values of $n$ is $6$

Therefore, the correct option is A.