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You know 1 is not a root because the sum of the coeffcients of the equation is 14, not zero.
It is fairly easy to try 3 by synthetic division (see attachment), which tells you that 3 is a root and the remaining quadratic factor is x²-3x-5. The quadratic formula tells you the roots of that factor are
... x = (-b±√(b²-4ac))/(2a) = (3±√29)/2
The appropriate choices are
... C. (3-√29)/2
... D. (3+√29)/2
... F. 3
_____
The quadratic formula tells you the solution to
... ax²+bx+c=0
is x = (-b±√(b²-4ac))/(2a)
We have a=1, b=-3, c=-5.
