The equation 3x+6=-6x^2 is equivalent to
6x^2+3x+6=0
In an equation ax^2+bx+c=0, the roots are given by x=(-b+-sqrt(b^2-4ac))/(2a)
It is apparent that root mainly depend on the nature of b^2-4ac, called the discriminant, though it also depends on the values that a,b and c can take.
Assuming that a,b,c in QQ or are just integers, the type of roots are decided by the discriminant Delta=b^2-4ac.
If Delta=0, the equation has just one root or two coincident (or identical) roots.
If Delta is a perfect square (of course Delta>0), roots are rational and can be easily determined by the usual method of splitting the middle term and factorizing ax^2+bx+c.
If Delta>0 but not a perfect square, roots are real.
If Delta<0, the roots are complex. Note that if along with Delta<0, we have b=0, the roots are imaginary.
In the given example, Delta=3^2-4xx6xx6=9-144=-135 and as b!=0, roots are complex.
and x=(-3+-sqrt(-135))/12=-1/4+-(3sqrt15)/4i
