The length of a rectangle is 4 less than twice

Let weigth of rectangle be w
length of rectangle be ${\displaystyle 2w-4}$
Area of rectangle =70 sq. feetask
${\displaystyle lenght\times weight=70}$
${\displaystyle (2w-4)w=70}$
${\displaystyle 2w^2-4w=70}$
${\displaystyle 2w^2-4w-70=0}$
${\displaystyle w^2-2w-35=0}$
${\displaystyle w^2-(7-5)w-35=0}$
${\displaystyle w^2-7w+5w-35=0}$
${\displaystyle w(w-7)+5(w-7)=0}$
${\displaystyle (w-7)(w+5)=0}$
${\displaystyle w=7,-5}$
${\displaystyle w=-5}$ is not possible [width cannot be negtive]
$\Rightarrow \text{width of rectangle}=7ft$
${\displaystyle \Rightarrow \text{Length of rectangle}=(2\times 7)-4}$
${\displaystyle =14-4}$
=10ft