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#
