Let's denote Sean's speed on the way to the concert as \( x \) miles per hour.
The time taken to travel to the concert can be expressed using the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
So, the time taken to get to the concert is:
\[ \text{Time to concert} = \frac{120}{x} \]
On the way back, Sean drove 10 mph slower, so his speed was \( x - 10 \) mph. The time taken to return home is then:
\[ \text{Time back home} = \frac{120}{x - 10} \]
According to the problem, the trip home took 2 hours longer than the trip to the concert, which gives us the equation:
\[ \frac{120}{x - 10} = \frac{120}{x} + 2 \]
To eliminate the fractions, we can multiply every term by \( x(x - 10) \):
\[ 120x = 120(x - 10) + 2x(x - 10) \]
Expanding both sides, we have:
\[ 120x = 120x - 1200 + 2x^2 - 20x \]
We can simplify this equation by moving all terms to one side:
\[ 0 = -1200 + 2x^2 - 20x \]
This can be rewritten as:
\[ 2x^2 - 20x - 1200 = 0 \]
Now, we can simplify it further by dividing the entire equation by 2:
\[ x^2 - 10x - 600 = 0 \]
Next, we solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -10 \), and \( c = -600 \):
\[ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-600)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (-10)^2 - 4 \cdot 1 \cdot (-600) = 100 + 2400 = 2500 \]
Now applying this back to the formula:
\[ x = \frac{10 \pm \sqrt{2500}}{2} = \frac{10 \pm 50}{2} \]
This gives us two potential solutions for \( x \):
- \( x = \frac{60}{2} = 30 \)
- \( x = \frac{-40}{2} = -20 \) (which we discard since speed cannot be negative)
So, the speed Sean drove on the way to the concert is:
\[ \boxed{30} \text{ miles per hour.} \]
To verify, if Sean drove at 30 mph to the concert, then the return trip speed was \( 30 - 10 = 20 \) mph.
The time taken for the trip to the concert is:
\[ \frac{120}{30} = 4 \text{ hours} \]
The time taken for the return trip is:
\[ \frac{120}{20} = 6 \text{ hours} \]
The difference in time is indeed:
\[ 6 - 4 = 2 \text{ hours} \]
Thus, the solution is confirmed, and Sean drove at \( \boxed{30} \) mph to the concert.
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