Question 965808
if a rectangular patio is {{{4yd}}} longer than it is wide, then {{{L=W+4yd}}}
{{{1yd=3ft}}}, so {{{L=W+12ft}}}
 
if the area is {{{A=448ft^2}}}, then

{{{L*W=448ft^2}}}....substitute {{{L}}}

{{{(W+12ft)*W=448ft^2}}}
 
{{{W^2+12ft*W=448ft^2}}}

{{{W^2+12ft*W-448ft^2=0}}}...factor completely

{{{W^2+28ft*W-16ft*W-448ft^2 = 0}}}......group

{{{(W^2+28ft*W)-(16ft*W+448ft^2 )= 0}}}

{{{W(W+28ft)-16ft(W+28ft)= 0}}}

{{{(W-16ft)(W+28ft) = 0}}}

solutions:

if 
{{{(W-16ft) = 0}}}=>{{{W=16ft}}}

if 
{{{(W+28ft) = 0}}}=>{{{W=-28ft}}}=>the width cannot be negative, so disregad this solution

so, the width in feet is {{{W=16ft}}}

and the length is {{{L=16ft+12ft}}}=>{{{L=28ft}}}