As #x^2+7x+6<=6#, we have #x^2+7x+cancel6<=cancel6# or
#x^2+7x<=0# or #x(x+7)<=0#
If product of #x# and #x+7# is negative (for the time we are ignoring equality sign - note that for equality we just add #x=0# and #x=-7# to the solution),
either #x>0# and #x+7<0# i.e. #x<-7# - but #x>0# and #x<-7# together is just not possible.
or #x<0# and #x+7>0#, i.e. #x>-7# - which is possible if #x# lies between #-7# and #0# i.e. #-7< x < 0#.
Now including the equality sign
Solution is #-7<= x <= 0#.
You may also see from the graph that #x^2+7x# is negative in the interval #-7<= x <= 0#.
graph{x(x+7) [-10, 5, -20, 20]}
