Two cars leave an intersection at the same time. one heads due south and the other heads due east. later, the two cars are exactly 50 mi apart. the car headed south has traveled 10 mi farther than the car going east.

how far has the car going east traveled?

Let O be the origine (or the departure point) and OA (south) and OB East
We have now a right triangle with the hypotenuse AB = 50 mi.
If OB = x, then OA = x +10 (given). Now apply Pythagoras:
OA² + OB² = AB²
(x+10)² + x² = 50²
x²+20x+100+x² = 2500
2x²+20x-2400 =0
Solve for x this quadratic equation and you will find x' = 30 and x" = -40
Obviously x" = -40 is an extraneous solution. So the answer is:
 OB = x= 30 mi (travelled EAST)
OA=x+10 = 40 mi (travelled south)

Let n be the car going east. Then the south-bound car is n+10. So:
n²+(n+10)²=50²
2n²+20n-2400=0
n²+10n-1200=0
(n+40)(n-30)=0
n=30 miles traveled by the car going east; 40 miles by car going south. ☺☺☺☺