Question 453088
Let x= the width and therefore length= 3+x
Area= L*W = (3+x)*x 
So by simple simplification, the area has the equation,{{{x^2+3x}}}
We are bounded to find values of width within the areas of 154 and 460. Which means this equation must hold true, {{{154<x^2+3x<460}}}
we solve two equations:
{{{x^2+3x<460}}}...... eqn 1
{{{x^2+3x>154}}} ..... eqn 2
The two equations have a quadratic system that is {{{ax^2+bx+c=0}}}
Eqn 1: {{{x^2+3x-460<0}}} using this formula {{{x<(-b+-sqrt(b^2-4*a*c))/(2*a)}}}
will give {{{x<-23 and x<20}}} that is: {{{x<(-3+-sqrt(3^2-4*1*-460))/(2*1)}}}

Eqn 2: {{{x^2+3x-154>0}}} also with the formula gives {{{x>-14 and x>11}}} that is: {{{x>(-3+-sqrt(3^2-4*1*-154))/(2*1)}}}

Note: The areas 154 and 460 are included and we deal with positive values because we are looking for measurements.

So we have 11 and 20 from eqn 2 and 1 respectively. However, the possible values  for the Width in order to find the areas between 154 and 460 ranges from 11 to 20.
Mathematically: {{{11<x<20}}} is the solution, 11 and 20 are included.