Question 1009910

The length {{{L}}} of a rectangle is {{{6}}}meters more than the width {{{W}}}. 

{{{L=W+6}}} 
The area is {{{A=55}}} square meters. 

{{{A=LW}}}

{{{55=(W+6)W}}}

{{{55=W^2+6W}}}

{{{W^2+6W-55=0}}}

{{{W^2-5W+11W-55=0}}}

{{{(W^2-5W)+(11W-55)=0}}}

{{{W(W-5)+11(W-5)=0}}}

{{{(W-5)(W+11) =0}}}

solutions:
{{{(W-5) =0}}}=>{{{W=5}}}
{{{(W+11) =0}}}=>{{{W=-11}}}

since we are looking for the width, we will need only positive solution
so, the width of the rectangle is {{{highlight(W=5)}}}meters

go to {{{L=W+6}}} and find the length 

{{{L=5+6}}}

{{{highlight(L=11)}}} meters ->the length  of the rectangle