Question 1077389

The area of a rectangle patio is {{{700}}} square feet. 

If the length {{{L}}} of the patio is {{{5}}} feet less than {{{twice}}} its width {{{W}}}, we have
 
{{{L=2W-5}}}

The area of a rectangle is: {{{A=L*W}}}

{{{700=(2W-5)*W}}}
{{{700=2W^2-5W}}}
{{{2W^2-5W-700=0}}}.........factor
{{{2W^2+35W-40W-700=0}}}
{{{(2W^2-40W)+(35W-700)=0}}}
{{{2W(W-20)+35(W-20)=0}}}
{{{(W - 20)(2W + 35) = 0}}}

solutions:
if {{{(W - 20) = 0}}}=>{{{W=20}}}
if {{{(2W + 35) = 0}}}=>{{{2W=-35}}}=>{{{W=-35/2}}}....disregard negative solution, width cannot be negative

so, {{{highlight(W=20)}}}

now find length
{{{L=2W-5}}}=>{{{L=2*20-5}}}=>{{{highlight(L=35)}}}

the dimensions of the patio:{{{highlight(L=35)}}} and {{{highlight(W=20)}}}