Make an equation system based on the problem
"The length of a rectangle is 3 cm less than twice the width" could be written as
∴ l = 2w - 3 (first equation)
"The area is 54" could be written as
∴ l × w = 54 (second equation)
Substitute (2w - 3) into l in the second equation
l × w = 54
(2w - 3) × w = 54
2w² - 3w = 54
2w² - 3w - 54 = 0
(2w + 9)(w - 6) = 0
The zeros of the equation
2w + 9 = 0 or w - 6 = 0
w = -9/2 or w = 6
Because the width couldn't be negative, -9/2 is excluded from the solution. The width of the rectangle is 6 cm
