Assume that [tex]p\ \textgreater \ 0[/tex] and multiply the equation [tex] \frac{1}{4p} (x-h)^2+k=0[/tex] by 4p. Then you obtain the equation [tex](x-h)^2+4pk=0[/tex].
1) If k>0, then 4pk>0 and the equation doesn't have solutions, because [tex](x-h)^2=-4pk<0[/tex] and this is unreal.
2) If k=0, then 4pk=0 and [tex](x-h)^2=0[/tex]. There is one solution x=h.
2) If k<0, then 4pk<0 and the equation
[tex](x-h)^2=-4pk>0[/tex] has two different solutions [tex]x_1=h+ \sqrt{-4pk} [/tex] and [tex]x_2=h- \sqrt{-4hk} [/tex].
