Amotorboat that travels with a speed of 20 km/hour in still water has traveled 20 km against the current and 180 km with the current, having spent 8 hours on the entire trip. find the speed of the current of the river

Let C=current speed
t=time in hours against the current
8-t= time going with the current
20+c=speed of the boat going with the current
20-c=speed of the boat going against the current
Distance=speed*time
thus;
20=(20-c)*t
Going with current
180=(20+c)*(8-t)
but
t=20/(20-c)
180=(20+c)(8-20/(20-c))
180=20(8-20/(20-c))+c(8-20/(20-c))
180=160-400/(20-c)+8c-20c/(20-c)
multiplying through by (20-c)
180(20-c)=160(20-c)-400+8c(20-c)-20c
3600-180c=3200-160c-400+160c-8c^2-20c
This will give us;
3600-3200-180c+400+8c^2+20c=0
800-160c+8c^2=0
dividing through by 8 we get:
100-20c+c^2=0
solving the above quadratic equation we get:
(c-10)^2
thus
c=10 km/h
The answer is 10 km/h

Ab=dc ac=de ad=bce i hope this !