Question 1118753
Looking at a few factorization trial combinations for the quadratic lead to
{{{(2x+4)(x+3)=2x^2+(-1)*5(n-1)x+12=00}}}
and these roots are  -3 and -2,  which are consecutive integers.



Steps from that and then compare to the given quadratic expression:
{{{(2x+4)(x+3)}}}
{{{2x^2+4x+6x+12}}}
{{{2x^2+10x+12}}}
from which corresponding parts give
{{{-5(n-1)=10}}}



Find n from that last equation.