Question 898976
Let y to be the length, while z to be the width.
length, {{{y=z+17}}}
width, {{{z=z}}}
{{{Area<60m^2}}}
{{{y*z<60}}}
{{{(z+17)*z<60}}}
{{{z^2+17z<60}}}
{{{z^2+17z-60<0}}}
{{{(z+20)(z-3)<0}}}
The range of z is {{{-20<x<3}}}, which holds the number -19,-18,-17,-16,-15,-14,-13,-12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2.
Since the length for an area cannot be zero and in negatives, the range of z are {{{z>0}}},{{{z<3}}}.

To get the possible lengths, substitute {{{z=1}}} and {{{z=2}}} into {{{y=z+17}}}.
{{{y=1+17}}}
{{{y=18}}}
or
{{{y=2+17}}}
{{{y=19}}}

Therefore, the possible lengths are 18m and 19m.

**In fact, the value z in range of {{{z>0}}},{{{z<3}}} do not only include 1 and 2, but also involve the positive rational numbers, such as 
0.00000...1, 0.001, 0.01, 0.1, 0.2, 0.26, 1.26895, 1.5, 1.55, 1.88888, 2.12, 2.563, 2.889, 2.9, 2.93, 2.99999999.
*As conclusion, value z can be any positive rational numbers as long as it is in this range: {{{z>0}}},{{{z<3}}}.