The equation $x^{2}+2kx+7k-10=0$ has two equal real roots precisely when the discriminant of this quadratic equation equals 0. The discriminant, $\Delta$, equals $\Delta=(2k)^{2}-4(1)(7k-10)=4k^{2}-28k+40$. For the discriminant to equal 0, we have $4k^{2}-28k+40=0$ or $k^{2}-7k+10=0$ or $(k-2)(k-5)=0$. Thus, $k=2$ or $k=5$. We check that each of these values gives an equation with the desired property. When $k=2$, the equation is $x^{2}+4x+4=0$ which is equivalent to $(x+2)^{2}=0$ and so only has one solution for $x$. When $k=5$, the equation is $x^{2}+10x+25=0$ which is equivalent to $(x+5)^{2}=0$ and so only has one solution for $x$. The sum of these values of $k$ is $2+5=7$.