To solve the problem, let's define the variables and set up the equations based on the information given.
Let:
- \( r \) = speed (in mph) at which Sean traveled to the concert.
- The distance to the concert is 120 miles.
The time taken to travel to the concert can be expressed as: \[ \text{Time to concert} = \frac{120}{r} \]
On the way back, Sean drove 10 mph slower, so his speed was \( r - 10 \) mph. The time taken for the return trip can be expressed as: \[ \text{Time back home} = \frac{120}{r - 10} \]
According to the problem, the trip home took 2 hours longer than the trip to the concert. Therefore, we set up the equation: \[ \frac{120}{r - 10} = \frac{120}{r} + 2 \]
Next, we will solve this equation step by step. First, we'll eliminate the fractions by multiplying every term by \( r(r - 10) \): \[ 120r = 120(r - 10) + 2r(r - 10) \]
Expanding both sides: \[ 120r = 120r - 1200 + 2r^2 - 20r \]
Now, we simplify the equation: \[ 0 = -1200 + 2r^2 - 20r \] \[ 2r^2 - 20r - 1200 = 0 \]
To simplify, divide the entire equation by 2: \[ r^2 - 10r - 600 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -10 \), and \( c = -600 \): \[ r = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-600)}}{2 \cdot 1} \] \[ r = \frac{10 \pm \sqrt{100 + 2400}}{2} \] \[ r = \frac{10 \pm \sqrt{2500}}{2} \] \[ r = \frac{10 \pm 50}{2} \]
This leads to two possible solutions: \[ r = \frac{60}{2} = 30 \quad \text{or} \quad r = \frac{-40}{2} = -20 \]
Since speed cannot be negative, we take \( r = 30 \) mph.
To confirm:
- Time to the concert: \( \frac{120}{30} = 4 \) hours.
- Speed on the way back: \( 30 - 10 = 20 \) mph.
- Time back home: \( \frac{120}{20} = 6 \) hours.
The difference in time is indeed \( 6 - 4 = 2 \) hours, confirming that our solution is correct.
Thus, the speed at which Sean drove to the concert is: \[ \boxed{30} \text{ mph} \]