The area of rectangle whose length is 5 more than twice its width is 75 sq units. What is the perimeter of the rectangle?

Let the width of the rectangle = k units
∴ Length = (2k + 5) units.
According to the question,
Area = k(2k + 5)
⇒ 75 = 2k² + 5k
⇒ 2k² + 5k - 75 = 0
⇒ 2k² + 15k - 10k - 75 = 0
⇒ k(2k + 15) - 5(2k + 15) = 0
⇒ (2k + 15) (k - 5) = 0
⇒ k = 5 and -15/2
Width cannot be negative.
∴ Width = 5 units
∴ Length = 2x + 5 = 2 x 5 + 5 = 15 unit
∴ perimeter of the rectangle = 2(15 + 5)
= 40 units