Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution. 
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.

My attempt is as follows:
$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\right)<0$$
$$\left(x-\dfrac{k-\sqrt{k^2-5}}{5}\right)\left(x-\dfrac{k+\sqrt{k^2-5}}{5}\right)<0$$
$$x\in\left(\dfrac{k-\sqrt{k^2-5}}{5},\dfrac{k+\sqrt{k^2-5}}{5}\right)$$
As it is given that it has got only one integral solution, so there must be exactly one integer between $\dfrac{k-\sqrt{k^2-5}}{5}$ and $\dfrac{k+\sqrt{k^2-5}}{5}$
Let $x_1=\dfrac{k-\sqrt{k^2-5}}{5}$ and $x_2=\dfrac{k+\sqrt{k^2-5}}{5}$ , then $[x_2]-[x_1]=1$ where [] is a greater integer function.
But from here, how to proceed? Please help me in this.

hint
Observe that
$$x_2=\frac{k+\sqrt{k^2-5}}{5}=\frac{1}{k-\sqrt{k^2-5}}$$
$ x_2 $ is integral if  $ (k -\sqrt{k^2-5})=\pm 1$ , 
By the same, $ x_1 $ is integral if
$ (k+\sqrt{k^2-5}) =\pm 1$.
This gives the possible values for $ k$ :
$-4, -3, 3, 4$.
the sum of positives values is $7$.
