Solve by factoring 2x^2+2kx-14x+6-k=0

Question

how do I solve this problem? I think this may be unfactorable.

Answer 1

Divide by 2, factorise and rearrange:

x²+x(k-7)=½(k-6),

x²+x(k-7)+¼(k-7)²=½(k-6)+¼(k-7)², (completing the square)

(x+½(k-7))²=½k-3+¼k²-7k/2+49/4,

x+½(k-7)=±√(¼k²-3k+37/4)=±½√(k²-12k+37),

x=-½(k-7)±½√(k²-12k+37)=-½(k-7)±½√((k-6)²+1).