Question 884403
x may be any one of three different quantities.  I will use w for width and L for length.  No use of x.


The first sentence becomes {{{L=1+2w}}}.
Area is {{{wL=w(2w+1)=80}}}.


{{{2w^2+w=80}}}
{{{2w^2+2w-80=0}}}
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Discriminant, {{{2^2-4*2*(-80)=4+8*80=644}}}
{{{sqrt(644)=sqrt(4*161)=sqrt(4*7*23)=2sqrt(161)}}}
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{{{w=(-2+- 2sqrt(161))/4}}}
You must use the plus square root in this case.
{{{w=(-2+2sqrt(161))/4}}}
{{{highlight(w=(-1+sqrt(161))/2)}}}


Use solution for w to find solution for L.
{{{L=1+2w}}}
{{{L=1+2(-1+sqrt(161))/2}}}
{{{L=1+(-1+sqrt(161))}}}
{{{highlight(L=sqrt(161))}}}


You can go further and and use Pythagorean Theorem to get the diagonal:
Let r = the diagonal,
{{{r^2=L^2+w^2}}}
{{{r^2=161+((-1+sqrt(161))/2)^2}}}
{{{r^2=161+((sqrt(161)-1)/2)^2}}}
{{{r^2=161+(sqrt(161)-1)^2/4}}}
{{{r^2=(644+(sqrt(161)-1)^2)/4}}}
{{{(644+161-2sqrt(161)+1)/4}}}

{{{(644+162-2sqrt(161))/4}}}

{{{(806-2sqrt(161))/4}}}

{{{r^2=(806-2sqrt(161))/4}}}

{{{r=sqrt((806-2sqrt(161))/4)}}}

{{{highlight(r=sqrt(806-2sqrt(161))/2)}}}, the diagonal across the garden.