Hello,earachefl!
A NorthSouth highway intersects an EastWest highway at a point $\displaystyle P$.
An automobile crosses $\displaystyle P$ at 10:00 AM, traveling east at a constant rate of 20 mph.
At that same instant another automobile is two miles north of $\displaystyle P$, traveling south at 50 mph.
(a) Find a formula which expresses the distance $\displaystyle d$ between the automobiles
. . .at time $\displaystyle t$ (hours) after 10:00 AM.
(b) At what time will the automobiles be 104 miles apart?
I don't agree with either of their answers . . . Code:
Q *

50t 

B *
 * d
250t  *
 *
  P *      *  
 20t A

At 10 AM, the first car is at point $\displaystyle P$.
. . $\displaystyle t$ hours later, it is at point $\displaystyle A\!:\;PA = 20t$
At 10 AM, the second car is at point $\displaystyle Q\!:\;QP = 2$.
. . $\displaystyle t$ hours later, it is at point $\displaystyle B\!:\;QB = 50t$.
. . Hence: .$\displaystyle BP = 250t$
Pythagorus says: .$\displaystyle d^2\:=\:(20t)^2 + (250t)^2$
(a) Then we have: .$\displaystyle \boxed{d \:=\:\sqrt{2900t^2  {\color{red}200}t + 4}}$
If $\displaystyle d = 104\!:\;\;\sqrt{2900t^2  200t + 4} \:=\:104$
Square both sides: .$\displaystyle 2900t^2  200t + 4 \:=\:10816\quad\Rightarrow\quad 2900t^2  200t  10812 \:=\:0$
Divide by 4: .$\displaystyle 725t^2  50t  2703 \:=\:0$
Quadratic Formula: .$\displaystyle t \;=\;\frac{(50) \pm\sqrt{(50)^2  4(725)(2703)}}{2(725)} $
. . $\displaystyle t \;=\;\frac{50\pm\sqrt{7841200}}{1450} \;=\;\frac{50 \pm 20\sqrt{19603}}{1450} \;=\;\frac{5 \pm 2\sqrt{19603}}{145} $
And the positive root is: .$\displaystyle \boxed{t \:=\:\frac{5 + 2\sqrt{19603}}{145} \:\approx\:1.966\text{ hours.}}$