From the question;
The length of a rectangle is 3 m less than double the width
[tex]\begin{gathered} \text{let width = w } \\ \text{therefore} \\ \text{length = (2w - 3)m} \end{gathered}[/tex]The are of the rectangle is 27 square meters
therefore
[tex]\begin{gathered} \text{Area = length }\times\text{ width} \\ \text{27 = (2w - 3)w} \\ 27=2w^2-3w^{} \\ 2w^2-3w^{}\text{ - 27 = 0} \\ by\text{ factorising} \\ 2w^2+6w-9w\text{ - 27 = 0} \\ 2w(w\text{ + 3) }-9(w\text{ + 3) = 0} \\ (2w-9)(w^{}\text{ + 3) = 0} \\ 2w-9=0^{}\text{ or w + 3 = 0} \\ w\text{ = }\frac{9}{2}\text{ or w}^{}\text{ = -3} \end{gathered}[/tex]Since width of a rectangle can not be negative then
the width of the rectangle = 9/2m
The lentgh will be
[tex]\begin{gathered} \text{Length of rectangle = (2w -3})m \\ \text{Length of rectangle = (2(}\frac{9}{2})\text{ - 3})m \\ \text{Length of rectangle = (9 - 3})m \\ \text{Length of rectangle = 6m} \end{gathered}[/tex]Therefore the length of the rectangle is 6m
