Distance /speed = time.

Let x-5 kms/hr be the original speed . If you increase the speed by 5 kms then the speed would be x kms  per hour .

Then the time  taken to traverse the distance at speeds x-5 and x is 300/(x-5) hr and 300/x hr respectively. But  at 5 kms higher speed ,i.e, x kms/ hr, the time taken is 300/x hrs is less than the  time 300/(x-5) hrs  at the speed (x-5) kms/hr.

Therefore, the required equation is :

300/(x-5) = 300/x +2.

Multiply by x(x-5),the LCM of the dinominators :

300x=300(x-5)+2x(x-5)

300x=300x-1500+2x^2-10x

2x^2-10x-1500=0

Divide both sides by 2:

x^2-5x-750=0

(x-5/2)^2-(5/2)^2 -750=0

(x-2.5)^2=756.25.

Take square root on both sides:

x-2.5= sqrt(756.25)= + or -27.5.

x=27.5+2.5=30 kms or x =2.5- 27.5=-25kms, which inot practical.

30 lms/hr is the speed 5 kms higher than the original speed.

Therefore the original speed = x-5 = 30-5=25 kms/hr

Tally:

300/(x-5) = 300/25=12 hrs

300/x =300/30=10 hrs , which is less by 2 hrs