Let's denote Sean's speed on the way to the concert as \( x \) mph. According to the problem, the trip to the concert, which is 120 miles away, can be expressed as:
\[ \text{Time to concert} = \frac{120}{x} \]
On the way back, his speed is \( x - 10 \) mph, and the time taken to return home is:
\[ \text{Time back home} = \frac{120}{x - 10} \]
We know that the return trip took 2 hours longer than the trip to the concert. Thus, we can set up the equation:
\[ \frac{120}{x - 10} = \frac{120}{x} + 2 \]
Now, let's solve this equation. First, we'll multiply both sides by \( x(x - 10) \) to eliminate the fractions:
\[ 120x = 120(x - 10) + 2x(x - 10) \]
Expanding both sides gives:
\[ 120x = 120x - 1200 + 2x^2 - 20x \]
Now, simplify the equation:
\[ 0 = -1200 + 2x^2 - 20x \]
Rearranging the equation:
\[ 2x^2 - 20x - 1200 = 0 \]
We can simplify this by dividing the entire equation by 2:
\[ x^2 - 10x - 600 = 0 \]
Next, we can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -10 \), and \( c = -600 \):
\[ b^2 - 4ac = (-10)^2 - 4(1)(-600) = 100 + 2400 = 2500 \]
So, we find:
\[ x = \frac{10 \pm \sqrt{2500}}{2} = \frac{10 \pm 50}{2} \]
This gives us two potential solutions:
\[ x = \frac{60}{2} = 30 \quad \text{and} \quad x = \frac{-40}{2} = -20 \]
Since speed cannot be negative, we take the positive solution:
\[ x = 30 \]
Thus, Sean drove 30 mph on the way to the concert.
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