jmc

The domain of the function is the set of all real numbers if and only if the denominator $-5x^2+2x+k$ is nonzero for all $x.$ In other words, the quadratic $$-5x^2+2x+k=0$$should not have any real solutions. This means that the discriminant is negative, i.e. $$4-4(-5)(k)=4+20k<0.$$Solving, we find $k<-\frac{1}{5}.$ Therefore, the set of all possible $k$ is $\boxed{\left(-\infty ,-\frac{1}{5}\right)}.$