Im confused with setting this problem up

Question

A house mover towed a historic Victorian home 72 miles to locate it on a new site. On his return, without the heavy house in tow, his average speed was 25 mph faster and the trip was 2 hours shorter. How fast did he drive in each direction

Answer 1

distance=rate*time

d=rt

72=rt

72=(r+25)(t-2)

rt=(r+25)(t-2)     (they both equal 72, so set them equal to each other)

rt=rt+25t-2r-50

0=25t-2r-50

if 72=rt, then r=72/t

substitute

0=25t-2(72/t)-50

multiply both sides by t

0=25t²-144-50t

25t²-50t-144=0

factor

25*144=5*5*2*2*2*2*3*3=(2*2*2*5)*(2*3*3*5)=40*90 and 90-40=50, the coefficient of the middle term

(5t     )(5t    )=0 so far...

(5t    8)(5t    18)=0 so far...

(5t+8)(5t-18)=0

5t-18=0

5t=18

t=18/5

t=3.6 hours

72=rt

72=r*3.6

720=r*36

r=720/36

r=20 mph to new site

r+25=20+25=45 mph returning