Question 521731
let x be the usual average speed and t the among of times she usually  took to complete the trip.  let s be the average speed , t' the among of time she takes now and d the distance.

 s= x + 10    since her average speed now is 10 miles per hour more than her usual speed
 d= 200 miles  
 t'= t - 1   since she completed the trip in 1 hour less time usual

we know that distance= speed x time 
so d= st'   and    d= xt

so      xt= 200
       (x+10)(t-1)=200

      xt=200
     xt - x + 10t -10= 200

   xt=200
  200 - x + 10t - 10 =200

  xt=200
  10t - x= 10
  
  x= 10(t - 1)
  xt=200

  x= 10(t-1)
  10(t-1)*t = 200

  10t^2 - 10t - 200 =0
 
  t^2 - t - 20 =0

let's call p(x)= t^2 - t - 20 and let's factorize this polynomial.
the determinant D= (-1)^2 - 4(1)(-20)= 81= (9)^2  
  t1= (1+9)/2 = 5
  t2= (1-9)/2 = -4
so the answers of the equation ( t^2-T-20=0) are 5 and -4, but since t is an among of time , t has to be a positive number, so answer can't be -4 , therefore the answer is 5. 

so t=5
x=10(t-1)
 = 10(5-1)
 = 40 

so, her usual driving speed is 40 miles per hour.