The area of a rectangle whose length is 5 more than twice its width is 75 square unit. What is the perimeter of the rectangle?

Let width $= w$

The length is 5 more than twice the width, so:

Length $= 2w + 5$

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The area of the rectangle is 75 square units.

Area $=$ Length $\times$ Width

$75 = (2w + 5) \times w$

$75 = 2w^2 + 5w$

$2w^2 + 5w - 75 = 0$

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Factoring the quadratic equation:

Looking for two numbers that multiply to $(2 \times -75) = -150$ and add to $5$.

These numbers are $15$ and $-10$.

$2w^2 + 15w - 10w - 75 = 0$

$w(2w + 15) - 5(2w + 15) = 0$

$(2w + 15)(w - 5) = 0$

$w = -7.5$ or $w = 5$

Since width cannot be negative, $w = 5$ units.

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Length $= 2w + 5$

Length $= 2(5) + 5$

Length $= 15$ units

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Perimeter of rectangle $= 2(\text{Length} + \text{Width})$

Perimeter $= 2(15 + 5)$

Perimeter $= 2(20)$

Perimeter $= 40$ units

Therefore, the perimeter of the rectangle is 40 units.