Question 109585
Are you trying to find a k that yields one real solution? If you are, then the discriminant must equal zero. So 


{{{b^2-4ac=0}}} which means {{{(2k)^2-4(k+3)(4)=0}}}



{{{(2k)^2-4(k+3)(4)=0}}}


{{{4k^2-16k-48=0}}}



*[invoke quadratic_formula 4, -16, -48, "k"]



So if k=-2 or k=6, then it will give us one solution for {{{(k+3)x^2 + 2kx + 4 }}}