Step-by-step explanation:
Let the width=x
Length =6x-4
As we know that in a rectangle
[tex]{\boxed{\sf Area=LB}}[/tex]
[tex]\hookrightarrow\sf x (6x-4)=20 [/tex]
[tex]\hookrightarrow\sf 6x^2-4x=20 [/tex]
[tex]\hookrightarrow\sf 6x^2-4x-20=0 [/tex]
[tex]\hookrightarrow\sf x=\dfrac {1}{3}+\dfrac {1}{3}\sqrt {31} \quad or \quad \dfrac {1}{3}+\dfrac {-1}{3}\sqrt {31}[/tex]
