SOLUTION: The area of a rectangle is 165 ft (2)squared. The length of the rectangle is 4 feet longer than the width. Find the length and width. I need to know the equation and how to solve.

Question 280270: The area of a rectangle is 165 ft (2)squared. The length of the rectangle is 4 feet longer than the width. Find the length and width. I need to know the equation and how to solve.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The area of a rectangle is 165 ft (2)squared. The length of the rectangle is 4 feet longer than the width. Find the length and width. I need to know the equation and how to solve.
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Let the width be "x":
Then the length is "x+4"
--------------
Area = x(x+4)
165 = x^2+4x
---
x^2 + 4x - 165 = 0
Quadratic formula:
x = [-4 +- sqrt(16-4*-165)]/2
x = [-4 +- sqrt(676)]/2
x = [-4 +- 26]/2
Positive solution:
x = [22/2]
x = 11 ft. (width)
x+4 = 15 ft (length)
===========================
Cheers,
Stan H.
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