The length of a rectangle is 2 ft less than three times the width. The area is 60 ft SQUARED. Find the length of the rectangle

w(3w-2) = 60

solve for w, then get length

To find the length of the rectangle, we need to set up an equation using the given information.

Let's start by assigning variables:

Let L be the length of the rectangle.
Let W be the width of the rectangle.

According to the problem, the length L is 2 feet less than three times the width W. We can represent this information as an equation:

L = 3W - 2

Additionally, we know that the area of the rectangle is 60 square feet. The area of a rectangle is given by the formula:

Area = Length × Width

Plugging in the values we have, we can write the equation:

60 = L × W

Now, we have a system of two equations:

L = 3W - 2
60 = L × W

To solve this system of equations, we can substitute the value of L from the first equation into the second equation:

60 = (3W - 2) × W

Expanding the equation:

60 = 3W^2 - 2W

Rearranging the equation to standard quadratic form:

3W^2 - 2W - 60 = 0

To solve a quadratic equation, we can factor, use the quadratic formula, or complete the square. In this case, let's solve it by factoring:

(3W + 10)(W - 6) = 0

Setting each factor equal to zero:

3W + 10 = 0 or W - 6 = 0

From the first equation:

3W = -10
W = -10/3

Since the width cannot be negative, we discard the negative solution. Therefore, the width of the rectangle is 6 feet.

Now, substitute the value of W back into one of the original equations to find the length L:

L = 3W - 2
L = 3(6) - 2
L = 18 - 2
L = 16

Therefore, the length of the rectangle is 16 feet.