Given:

Car travel for 11 hours.

First 100 km, he traveled at a certain speed,

Remaining 280km distance traveled by the increased speed by 15km/hr

Formula Used:

Speed = Distance/Time

ax² + bx + c = 0,

To solve the quadratic equation for bigger numbers,

Use for x = {b +/- √(b²  – 4ac)}/2a

Calculation:

Time = Distance/Speed

A car traveled for 11 hours.

Let the initial speed of a car be x

⇒ 11 = 100/x + 280/(x + 15)

⇒ 11 = {100(x + 15) + 280x}/x(x + 15)

⇒ 11 = {100x + 1500 + 280x}/x² + 15x

⇒ 11(x² + 15x) = 380x + 1500 

⇒ 11x2 + 165x = 380x + 1500

⇒ 11x2 + 165x – 380x – 1500 = 0

⇒ 11x2 – 215x – 1500 = 0

Compairing with the quadratic equation ax2 + bx + c = 0,

we get,

a = 11, b = – 215, c = –1500

Using formula {–b +/- √(b2  – 4ac)}/2a

⇒ {–(– 215) +/- √(– 2152  – 4(11)(–1500))}/2(11)

⇒ {215 +/- √(46225 + 66,000)}/22

⇒ {215 +/- √1,12,225}/22

⇒ (215 +/- 335)/22

⇒ (215 + 335)/22 OR (215 – 335)/22

⇒ 550/22 OR –120/22

⇒ 25 OR –5.4545

Speed can not be negative

Initial speed of a car is 25km/hr.

Speed = Distance/Time

⇒ 25 = 100/Time

⇒ Time = 100/25

⇒ Time = 4 hours.

∴ Time taken by him to cover 100km distance is 4 hours.