I was sitting down doing some math homework the other night, and I came up with a pretty clever method to solve some of the problems.  What I was working with was building an equation with given zeroes.  For some of them, like 5, -5, it's not so hard at all.  (x-5)(x+5)=0 is the difference of two squares, x²-25=0.  But what about -5+2i?  Well, imaginary roots always come in conjugate pairs, soooooo...(x-(-5+2i))(x-(-5-2i))=0? 
x²-(-5-2i)x-(-5+2i)x+(-5+2i)(-5-2i)=0   FOIL
x²+5x+2ix+5x-2ix+25+10i-10i-4i²=0  distribute and FOIL
x²+10x+25-4i²=0                                   combine terms
x²+10x+25-4(-1) =0                               define i²
x²+10x+25+4=0                                      multiply -4*-1
x²+10x+29=0                                           simplify     

Yeah, that gets really messy when you try to FOIL it out.  Instead, since I was solving so many quadratics by completing the square, I figured I'd just do it backwards...
x=-5±2i                                                       given solution
x+5=±2i                                                      add five to both sides
(x+5)²=-4                                                    square both sides (we have two solutions to incorporate)
x²+10x+25=-4                                            FOIL
x²+10x+29=0                                              Add four to both sides

I'm sure I'm not the first person to ever figure this out, but I figured it out without any help from anyone else, so I'm proud of it.  Even more proud than when that best acne product did what it was supposed to.  I even went back to other problems I'd already done, just to see it in action some more.  Of course, this works for all conjugate pairs, even that ±5 up there, although that one is simpler to do by recognizing the form, and irrational roots, like 1±sqroot7.  If you like this, try it out and post in the comments.