Answer:
The equation is:
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While, the value of
is:

Explanation:
Step 1: Creating the equation
The area for a rectangle is given by:
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The
is given as
, and
is equal to
, and
is equal to
.
Substitute these values into the formula:

Simplify the equation:

Step 2: Find the value of x
We have the equation as:
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To find the value of
, we can make use of the quadratic formula:

From the equation, we have:

Substitute these values into the formula:

Calculate the values:

Since using a negative value of
, can give negative lengths or widths (which is impossible), we will omit it.
Thus, the value of
is:
