A rectangle area spans . The length must be 20ft more than the width. What are the dimensions?

Spans what?

the blala

To find the dimensions of the rectangle, we'll use the information given: the length of the rectangle must be 20ft more than the width.

Let's represent the width of the rectangle as "x" ft. According to the given information, the length can be represented as "x + 20" ft.

The formula for the area of a rectangle is length multiplied by width, so we can write the equation as follows:

Area = Length * Width

Since the area is not provided in the question, we can't solve the equation directly. However, we can still find the dimensions by setting up the equation.

Let's assume the area of the rectangle is A. So the equation becomes:

A = (x + 20) * x

Now, we have an equation with one variable. Let's solve it to find the value of x, which represents the width.

Expanding the equation, we have:

A = x^2 + 20x

Rearranging the equation, we get:

x^2 + 20x - A = 0

This is a quadratic equation in the form of ax^2 + bx + c = 0, where a=1, b=20, and c=-A.

To find the value for x, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values, we get:

x = (-20 ± √(20^2 - 4(1)(-A))) / 2(1)

Simplifying further:

x = (-20 ± √(400 + 4A)) / 2

Now we have two possible values for x, represented by (-20 + √(400 + 4A))/2 and (-20 - √(400 + 4A))/2.

Since x represents the width, it cannot be negative, so we'll disregard the negative solution.

Therefore, the width of the rectangle is given by:

Width = x = (-20 + √(400 + 4A)) / 2

Now, we can substitute this value back into the equation for the length to find its value.

Length = Width + 20 = [(-20 + √(400 + 4A)) / 2] + 20

With this equation, we can calculate the dimension of the rectangle for any given area A.