Roots of $z^4 + 2z^3 + 20z + 12 = 0$ 
Find the roots of $$z^4 + 2z^3 + 20z + 12 = 0.$$

I still haven't been able to solve it after hours of trying. There are no rational roots and factoring seems impossible. Who of you, mathematical geniuses, can help me? :-)  

The rational roots test fail so there is no linear real factor but there may be some $(z^2 + az + b)(z^2 + cz + d)$ where
$a+c = 2$ and $b+ ac + d =0$ and $ad+bc = 20$ and $bd=12$
which .... isn't obvious but.... oh heck.  If we assume $\{b,d\} = \{\pm 1,12\}, \{\pm 2,6\}, \{\pm 3,4\}$ then $ac =-b-d= \mp 13, \mp 8, \mp 7$.  But  $a+c = 2$ so (assuming integers) $a$ and $c$ are opposite signs.  $13$ and $7$ are primes larger than $3$ so a difference of $2$ between their factors is not possible.  SO the nicest is $\{b,d\} = \{2,6\}$ and $ac = -8$.  And as $a+c = 2$ so $a=4; c=-2$.  (That's the nicest; it's not a guarentee.)
So $ad+bc= 4*(2,6)-2*(6,2) = 20$ is solved with $d=6; b= 2$.
Hurray!  $(z^2 +4z +2)(z^2 -2z + 6)=z^4 + 2z^3 + 20z +12$
And we can use quadratic formula to solve:
$z = -2 \pm \sqrt 2$ and $z= 1 \pm \sqrt{5}i$.
