x12+ax2+bx+c=0

So far I have found only following reducible solvable quadrinomials of 12th degree:

a) x12+19683x2+78732x+78732 = x12+39(x+2)2 = (x4+3x3+9x2+27x+27)(x8-3x7+54x4-81x3-243x2+2916)

b) x12+2993784300000x2+84440070000000x+595410750000000 = x12+253955(39x+550)2 = (x4-90x2+5400x+59400)(x8+90x6-5400x5-51300x4-972000x3+19197000x2+510300000x+10023750000)

where both octic factors are solvable because they have Galois group 8T47.

In case removed the constraint for x2 and x1  T. Piezas found infinite many of solvable and irreducible quadrinomials:  x12−3abcx5+a3cx3+b3c2  - see https://mathoverflow.net/q/443679

In case additional x3 so far found following irreducible and solvable examples - all of group 12T294:

a) x12+73693152x3-233361648x2+1400169888x-8526675600

b) x12+87864912x3+415704960x2-1774304352x-8717521968

c) x12+128887200x3+614109600x2+20879726400x+97816636800

d) x12+3133323648x3+70941940992x2+380846209536x+900416943360