Otto's rectangular garden measures 12 feet by 20 feet. He

Step 1
Given dimensions of rectangular garden is 12 feet and 20 feet.
Let uniform width of walkway be x feet.
Step 2
Area of garden $=20\times 12$ sq. feet=240sq. feetask
Total area of walkway and garden together is 660 sq. feet.
Therefore, area of walkway ${\displaystyle =\text{total area}-\text{area of garden}}$
=660-240 sq. feetask
=420 sq. feetask
From the figure we can find that,
Area of walkway ${\displaystyle =4x^2+2(20\cdot x)+2(12\cdot x)}$
Therefore ${\displaystyle 4x^2+2(20\cdot x)+2(12\cdot x)=420}$
${\displaystyle 4x^2+40x+24x=420}$
${\displaystyle 4x^2+64x-420=0}$
${\displaystyle 4(x^2+16x-105)=0}$
${\displaystyle x^2+16x-105=0}$
${\displaystyle x^2+21x-5x-105=0}$
${\displaystyle x(x+21)-5(x+21)=0}$
${\displaystyle (x+21)(x-5)=0}$
${\displaystyle x=-21\text{or}x=5}$
Step 3
Width of walkway can't be negative.
Therefore, ${\displaystyle x=5}$
Hence, width of walkway ${\displaystyle =5}$ feet

Length of the rectangular garden ${\displaystyle =20}$ feet
and width =12 feet
Area of the garden $=20\times 12$ sq. feetask
=240 sq. feetask
Suppose, the width of the walkway the garden ${\displaystyle =x}$ feet
Length of the garden walkway ${\displaystyle =(20+2x)}$ feet and width ${\displaystyle =(12+2x)}$ feet
Area of the garden with walkway
${\displaystyle =(20+2x)(12+2x)}$ sq. feet
${\displaystyle =(240+24x+40x+4x^2)}$ sq. feet
${\displaystyle =(4x^2+64x+240)}$ sq.feet
According to the given condition,
${\displaystyle 4x^2+64x+240=660}$
${\displaystyle 4x^2+64x+240-660=0}$
${\displaystyle 4x^2+64x-420=0}$
${\displaystyle 4(x^2+16x-105)=0}$
${\displaystyle x^2+16x-105=0}$
${\displaystyle x^2+21x-5x-105=0}$
${\displaystyle x(x+21)-5(x+21)=0}$
${\displaystyle (x+21)(x-5)=0}$
${\displaystyle (x+21)=0(x-5)=0}$
${\displaystyle x+21=0,thenx=-21}$
${\displaystyle x-5=0thenx=5}$
${\displaystyle x\ne -21\therefore x=5}$
width of the walkway =5 feet

Step 1
Given:
$=660-240\text{sq.feet}$
$=420\text{sq.feet}$
Area of walkway:
$4x^2+2(20\times x)+2(12\times x)$
So,
$4x^2+2(20\times x)+2(12\times x)=420$
Do the multiplications.
$4x^2+40x+24x=420$
Combine 40x and 24x to get 64x.
$4x^2+64x=420$
Subtract 420 from both sides.
$4x^2+64-420=0$
Divide both sides by 4.
$x^2+16x-105=0$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as
$x^2+ax+bx-105$
To find a and b, set up a system to be solved.
a+b=16
ab=1(-105)=-105
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -105.
-1,105
-3,35
-5,21
-7,15
Calculate the sum for each pair.
-1+105=104
-3+35=32
-5+21=16
-7+15=8
The solution is the pair that gives sum 16.
a=-5
b=21
Rewrite $x^2+16x-105$ as $(x^2-5x)+(21x-105)$
$(x^2-5x)+(21x-105)$
Factor out x in the first and 21 in the second group.
x(x-5)+21(x-5)
Factor out common term x-5 by using distributive property.
(x-5)(x+21)
To find equation solutions, solve x-5=0 and x+21=0
x=5
x=-21