For both solutions to have the same sign the constant in both factors have to have the same sign
which means that the last term of the quadratic must also be positive. If the last term is negative then the expression would factor to (x+a)(x-b) which eliminates #1,3and4
Equation #2 has no real solutions ( discriminators is <0) and so solutions will be in the form of a +bi, a -bi
where i is square root if (-1)
So that leaves only #5