Two cars left the city for a suburb, 120 km away, at the same time. The speed of one of the cars was 20 km/hour greater than the speed of the other, and that is why it arrived at the suburb 1 hour before the other car. What is the speed of both cars.?

Let #x=#hours it took the slower car to reach the suburb
The time it took the faster car to reach the suburb is #x-1#

Since distance#div#time#=#speed,
The speed of the slower car is #120"km"div(x)"h"=120/x"km/h"#
The speed of the faster car is #120"km"div(x+1)"h"=120/(x-1)"km/h"#

Since the faster car travelled #20"km/h"# faster than the slower car,
#120/(x-1)=120/x+20#

Now multiply both sides by #(x-1)# and then #x# to start solving.
#(120(x-1))/(x-1)=(120(x-1))/x+20(x-1)#
#(120x(x-1))/(x-1)=(120x(x-1))/x+20x(x-1)#

Simplify, distribute, factor, disregard, etc. Essentially, solve.
#(120xcolor(red)cancelcolor(black)(x-1))/color(red)cancelcolor(black)(x-1)=(120color(red)cancelcolor(black)x(x-1))/color(red)cancelcolor(black)x+20x(x-1)#
#color(red)cancelcolor(black)(120x)=color(red)cancelcolor(black)(120x)-120+20x^2-20x#
#0=20x^2-20x-120#
#0=x^2-x-6#
#0=x^2+2x-3x-6#
#0=x(x+2)-3(x+2)#
#0=(x+2)(x-3)#
#x+2=0orx-3=0#
#x="-"2orx=3#

Since #x# is the number of hours for the slower car to reach the suburb, a negative solution can be disregarded as extraneous as negative time does not (yet?) exist. #x=3#

We know that the speed of the slower car is #120/x"km/h"# and that of the faster car is #120/(x-1)"km/h"# so plugging in #x#, we get:
Slower car: #120/((3))=120/3=40"km/h"#
Faster car: #120/((3)-1)=120/2=60"km/h"#