x14+ax13+bx+c=0

Beside the trivial cases of the form xⁿ+ax^n-k+bx^k+ab=(x^k+a)(x^n-k+b)=0 (where n and k are coprime) so far I have found only one reducible and solvable quadrinomial of 14th degree: x¹⁴+(3/2)x¹³+(3/2)x+1=(x⁶+x⁵+x³+x+1)(2x⁸+x⁷-x⁶-x⁵-x³-x²+x+2)/2=0. Let us note that both factors (of the 6th and 8th degree) can be easily solved by substituting y=x+1/x.