# The lengthof a rectangle is 2 m more than its width. The area of the rectangle is 80 m^2. What is the length and the width of the rectangle?

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Let the length be L. Now length is 2 more than width, so width is equal to L - 2.

Now we have the area as 80 m^2

Area = length * width = L ( L - 2) = 80

=> L^2 - 2L = 80

=> L^2 - 10L + 8L - 80 = 0

=> L(L - 10) + 8( L - 10) = 0

=> ( L + 8) ( L - 10 ) =0

So L can be 10 or -8.

But as length is not negative we eliminate -8.

**Therefore the length is 10 m and width is 8 m.**

We'll note L the length and W the width.

We'll write mathematically the first sentence:

L = 2 + W

The area of the rectangle is:

A = Length*Width

From enunciation, we know that the area is A = 80 m^2.

80 = (2 + W)*W

We'll remove the brackets:

80 = 2W + W^2

We'll use symmetric property:

W^2 + 2W - 80 = 0

We'll apply the quadratic formula:

W1 = [-2 + sqrt(4 + 320)]/2

W1 = (-2 + 18)/2

W1 = 8

W2 = (-2-18)/2

W2 = -10

**Since it is about a length, we'll reject the negative value. We'll keep the positive value for W = 8 m.**

The length is:

L = 2 + W

L = 2 + 8

**L = 10 m**

The length of the rectangle is 2 feet more than the width.

So we presume the width to be x and length x+2.

Therefore the area of the rectangle = x(x+2) which is actually given to be 80m^2.

=> x(x+2) = 80.

= x^2+2x-80 = 0.

=> (x+10)(x-8) = 0.

=> x+10 = 0. Or x-8 = 0.

=> x= -10. Or x = 8.

So width = x= **8 feet**, length = x+2 = 8+2 = **10 feet**.