Question 1053736
Another way:
{{{(a-20)*(a+6)=5760}}} 
{{{5760=2^7*3^2*5}}} so {{{5760}}} has
{{{(7+1)*(2+1)*(1+1)=8*3*2=48}}} factors,
forming {{{48/2÷24}}} pairs.
We could list them all, starting with
{{{1*5760=5760}}}
{{{2*2880=5760}}}
{{{3*1930=5760}}} and so on.
However, we are looking for a
{{{(a-20)*(a+6)}}} product
of two factors that are
{{{(a+6)-(a-20)=26}}} units apart.
Our product would be towards the end of the list.
Since {{{5760}}} is between
{{{75^2=5625}}} and {{{76^2=5776}}} ,
and neither {{{75}}} nor {{{76}}}
is a factor 76of {{{5760}}} ,
we are looking for factors
{{{a-20<75}}} and {{{a+6>76}}} .
Obviously, 77, 78, and 79
are not factors of 5760,
but {{{80}}} is.
Likewise, it is easy to see that
81, 82, ....88, and 89 are not factors,
but {{{90}}} is.
That gives us two pairs:
{{{5760/80=72}}} so {{{72*80=5760}}}
would be the last pair in the list,
and {{{5760/90=64}}} so {{{64*90=5760}}}
is the second to last pair.
It is also the pair we want, because
{{{96-64=26}}} .
So, {{{a-20=64}}} ---> {{{a=64+20=highlight(84)}}} .