Move the #-41# to the left side by adding #41# to both sides:
#b^2-16b+41=cancel(-41+41)#
#b^2-16b+41=0#
We can then use the quadratic formula: # x = (-b \pm sqrt(b^2-4ac)) / (2a) #
where #a=1#, #b=-16#, and #c=41#
Substituting these values into the quadratic formula...
# x = (-(-16) \pm sqrt((-16)^2-4(1)(41))) / (2(1)) #
# x = (16 \pm sqrt(256-164))/2#
# x = (16 \pm sqrt(92))/2#
#---------------------#
#sqrt(92)# can be broken down into #sqrt(4)sqrt(23)# which simplifies to #2sqrt(23)#
#---------------------#
We now have,
# x = (16 \pm 2sqrt(23))/2#
# x = 8 \pm sqrt(23)#
Our final answer is then #x# or #b# in this case is equal to
#8+sqrt(23)#, #8-sqrt(23)#
