On a recent trip, sarah's car traveled 20 mph faster on the first 110 miles than it did on the remaining 80 miles. the total time for the trip was 4 hr. find the speed of sarah's car on the first part of the trip.

Let x mph be the speed for the 110 miles trip. Then for 80 miles,
Speed = (x-20) mph

Time, t = Distance/Speed
On 110 miles, t1 = 110/x hrs
On 80 miles, t2 = 80/(x-20) hrs

Total time = 4 hrs = t1+t2 = 110/x + 80/(x-20)

Solving for x;
4 = [110(x-20) + 80(x)]/x(x-20)
4(x)(x-20) = 110x -2200 + 80x
4x^2 - 80x = 110x - 2200 + 80x
4x^2 -80x -110x  - 80x +2200 = 0
4x^2 -270x +2200 = 0
Solving the quadratic equation;
x = [-(-270)+/- Sqrt ((-270)^2 -4(4)(2200)]/2*4 = 33.75+/- 24.27 = 9.48 mph or 58.02 mph

Ignore the value smaller than 20 as this would yield a negative value of second speed which is not practical.

Therefore, speed in the first 110 miles is 58.02 mph

Yes, that is the correct formula. to solve this problem given the number of years, plug in the number of years for t and plug the whole thing into the