We should accept x km/h to be the first speed of the vehicle.
We realize that,
Time = Distance/Speed
From the inquiry,
The time taken by the vehicle to finish 400 km = \[400/x\text{ }hrs\]
Presently, when the speed is expanded by 12 km.
Increased speed \[=\text{ }\left( x\text{ }+\text{ }12 \right)\text{ }km/h\]
What’s more, the new time taken by the vehicle to finish 400 km = \[400/\left( x\text{ }+\text{ }12 \right)\text{ }hrs\]
Along these lines, as per the inquiry we can compose
\[4800\text{ }x\text{ }3\text{ }=\text{ }5x\left( x\text{ }+\text{ }12 \right)\]
\[5×2\text{ }+\text{ }60x\text{ }\text{ }14400\text{ }=\text{ }0\]
Isolating by 5 we get,
\[x2\text{ }+\text{ }12x\text{ }\text{ }2880\text{ }=\text{ }0\]
\[x2\text{ }+\text{ }60x\text{ }\text{ }48x\text{ }\text{ }2880\text{ }=\text{ }0\]
\[x\left( x\text{ }+\text{ }60 \right)\text{ }\text{ }48\left( x\text{ }+\text{ }60 \right)\text{ }=\text{ }0\]
\[\left( x\text{ }+\text{ }60 \right)\text{ }\left( x\text{ }\text{ }48 \right)\text{ }=\text{ }0\]
In this way, \[x\text{ }+\text{ }60\text{ }or\text{ }x\text{ }\text{ }48\]
\[x\text{ }=\text{ }-\text{ }60\text{ }or\text{ }48\]
As, speed can’t be negative.
\[x\text{ }=\text{ }48\] is just substantial
Accordingly, the speed of the vehicle is \[48\text{ }km/h.\]