Answer:
Step-by-step explanation:
You want to know the width and length of a rectangle with an area of 54 square units when the length is 3 more than twice the width.
The formula for the area of a rectangle is ...
A = LW
The problem statement tells us L = 2W+3, so the given values let us write the equation ...
54 = (2W +3)W
Rewriting this a bit, we have ...
2W² +3W -54 = 0
We can factor this if we find two integers whose product is 2(-54) = -108, and whose sum is +3. Those are 12 and -9.
(2W -9)(2W +12)/2 = 0
(2W -9)(W +6) = 0
Solutions are values of W that make the factors zero:
2W -9 = 0 ⇒ W = 9/2 = 4.5
W +6 = 0 ⇒ W = -6
The width cannot be negative, so the solution is ...
W = 4.5
L = 2(4.5) +3 = 12
The rectangle is 4.5 units wide and 12 units long.