Let the slower speed be #x# mph.

The faster speed is therefore #x+10# mph

You know the distance and the speeds, write expressions for the each time..

#t = d/s#

#t_1 = 280/x# (longer time) #" "and t_2 = 280/(x+10)# (shorter time)

The difference between the two times was #1 " hour"#

#t_1 - t_2 = 1#

#280/x - 280/(x+10) = 1" "larr xx LCM# to cancel the denominators

#(color(blue)(x(x+10)xx280))/x - (color(blue)(x(x+10)xx280))/(x+10) = color(blue)(x(x+10)xx)1#

#280(x+10) -280x = x(x+10)#

#280x+2800 -280x = x^2+10x" "larr#

Now you have a quadratic equation, make it equal to 0.

#0 = x^2 +10x -2800#

This expression does not have factors.

Completing the square gives:

#x^2 +10x +25 = 2800+25#

#(x+5)^2 = 2825#

#x +5 = sqrt2825" "# only the positive root is valid

#x = sqrt2825 -5#

#x = 48.1507# mph - this is the slower speed

#x +10 = 58.1507# mph -this is the faster speed

#t_1 = 280/48.1507 = 5.815 hrs#

#t_2 = 280/58.1507 = 4.815 hrs#

#"ave speed" = ("total distance")/("total time")#

#"ave speed" = 560/10.63#

#"ave speed" = 52.68#mph

An expression for the ave speed would be

#"ave speed" = 560/(t_1+t_2)#