#(-5)/11=-0.45bar(45)#

#color(white)("XXX")#see below for solution using long division

This is simpler to see if we do the long division with positive numbers and then negate the quotient. (It is possible to do it with a negative dividend but the process tends to be ore error prone).

Set-up:
#color(white)("XXX")underline(color(white)("XXXXXXXXXX"))#
#11 color(white)("X")) color(white)("X")5.0 color(white)("X")0 color(white)("X")0color(white)("X") 0 color(white)("X")...#

#11# "goes into" #5#, #0# times
Subtract #0xx11# from the dividend
and bring down the next digit of the dividend (#0#)
#color(white)("XXX")underline(color(white)("XXXXXXXXX"))#
#11 color(white)("X")) color(white)("X")5.0 color(white)("X")0 color(white)("X")0color(white)("X") 0 color(white)("X")...#
#color(white)("XXXX")underline(0)#
#color(white)("XXXX")5color(white)(".")0#

#11# "goes into #50#, #4# times
Subtract #11xx4#
and "bring down" the next #0#
#color(white)("XXX")underline(color(white)("X")0.4 color(white)("XXXXXXX"))#
#11 color(white)("X")) color(white)("X")5.0 color(white)("X")0 color(white)("X")0color(white)("X") 0 color(white)("X")...#
#color(white)("XXXX")underline(0)#
#color(white)("XXXX")5color(white)(".")0#
#color(white)("XXXX")underline(4color(white)(".")4)#
#color(white)("XXXX4.")6color(white)("X")0#

Repeat this process:
#color(white)("XXX")underline(color(white)("X")0.4 color(white)("X")5color(white)("XXXXX))#
#11 color(white)("X")) color(white)("X")5.0 color(white)("X")0 color(white)("X")0color(white)("X") 0 color(white)("X")...#
#color(white)("XXXX")underline(0)#
#color(white)("XXXX")5color(white)(".")0#
#color(white)("XXXX")underline(4color(white)(".")4)#
#color(white)("XXXX4.")6color(white)("X")0#
#color(white)("XXXX4.")underline(5color(white)("X")5)#
#color(white)("XXXX4.5"X")5#

Notice that we now have a remainder of #5# just as we did after the first subtraction.
#rarr# the quotient pattern will repeat from this point on...
#color(white)("XXX")underline(color(white)("X")0.4 color(white)("X")5color(white)("X") bar(4 color(white)("X")5))#
#11 color(white)("X")) color(white)("X")5.0 color(white)("X")0 color(white)("X")0color(white)("X") 0 color(white)("X")...#

As noted at the beginning
if #5/11=0.45bar(45)#
then #(-5)/11=-0.45bar(45)#