Question 886874
Assigning variables,
Length is L, width is w.

{{{L=1+3w}}} and {{{wL=80}}}.  Substitution for L gives {{{w(3w+1)=80}}},  {{{3w^2+w=80}}}
{{{highlight_green(3w^2+w-80=0)}}}
You can try to factor that.  Product -80 will have various combinations but you need to also have
the coefficient of 3 on the leading term.
For 80:   20 & 4, 2 & 40, 8 & 10, ... others.
Easiest to avoid trying to factorize.
-
COMPLETE THE SQUARE-----
First factor the 3.
{{{3(w^2+(1/3)w-80/3)=0}}}
The term to use is {{{(1/6)^2}}}   (Reading-up on this will help);
{{{3(w^2+(1/3)w+1/36-1/36-80/3)=0}}}
{{{3((w+1/6)^2-(1/36+80/3))=0}}}
{{{3((w+1/6)^2-(961/36))=0}}}
{{{3(w+1/6)^2-961/12=0}}}
{{{3(w+1/6)^2=961/12}}}
{{{(w+1/6)^2=961/36}}}
{{{w+1/6=0+- sqrt(961/36)}}}
{{{w=-1/6+- 31/6}}}
{{{w=(-1-31)/6}}} OR {{{w=(-1+31)/6}}}
{{{cross(w=-32/6)}}}  OR  {{{highlight(w=5)}}}
The meaningful result here is w=5 for width.


You could have found 5 and 16 for your attempt at factorization of the general quadratic in w.
That seemed like something which would take too many different trys to find.


Now, you can just use the formula for L and the now known value for w.
{{{L=3w+1}}}
{{{L=3*5+1}}}
{{{highlight(L=16)}}}


Also, once getting the earlier simpler quadratic in w, the general solution could have been chosen:

{{{3w^2+w-80=0}}};
{{{w=(-1+- sqrt(1-4*3*(-80)))/(2*3)}}}
and then evaluate (and choose the value which makes sense.)