# The length of a rectangle is 2m more than the width. The area of the rectangle is 20m^2. Find the dimensions of the rectangle, to the nearest tenth.please explain in detail

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Let x be the width of the rectangle.

`width = x`

Then, express in math form the condition "length is 2 more than x".

`l e n g th = 2 + x`

Next, apply the formula for the area of rectangle.

`A = lw`

where A is the area, l is the length, and w is the width.

From here, substitute A=20. Also replace w with x and l with 2+x.

`20 = x(2+x)`

`20=2x+x^2`

Express the equation in quadratic form `ax^2+bx+c=0` .

`x^2+2x-20=0`

Use quadratic equation to solve for x.

`x=[-b+-sqrt(b^2-4ac)]/[2a] = [-2+-sqrt(2^2-4(1)(-20))]/(2*1)=(-2+-sqrt84)/2`

`x=(-2+-2sqrt21)/2 = -1+-sqrt21`

`x_1=-1+sqrt21=3.6 ` and `x_2=-1-sqrt21=-5.6`

Since x is the width of the rectangle, take only the positive value. Then, substitute this to determine the length.

`width=x=3.6` , `l e n g th = 2 + x = 2+3.6 = 5.6`

**Hence, the width of the rectangle is 3.6m and its length is 5.6m.**