Question 26966
On going he ws travelling at x mph 
On way back he was travelling at x-10mph.
Total distance = 80
20 minutes = 1/3h


{{{80/x+80/(x-10)=1/3}}}
{{{80((x)-(x-10))=1(x)(x-10)/3}}}
{{{800=(x^2-10x)/3}}}
{{{2400=x^2-10x}}}
{{{x^2-10x-2400=0}}}
Use quadratic formula and solve you get 
{{{x=5sqrt(97+1)}}}
x=54.2 (rounded)


54.2-10=44.2


Hence, he speed going was 54.2mph and his speed returning was 44.2mph.
Paul.