A well is 20ft deep. A snail at the bottom of the well climbs upto 4ft each day and slips back 2ft every night. how many days the snail to reach the too

Defining Variables:

Let the number of days the snail takes to climb = x days

Total distance climbed in x days = x*4

Total distance slipped in x days = 2*x

Solving for the number of days:

Since the total distance the snail climbs is 20ft (depth of well)

The total distance climbed - total distance slipped = 20

replacing the variables

4x - 2x = 20

2x = 20

x = 10

Therefore, it will take the snail 10 days to reach the top

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol