Please help

I get a=44  and  b =13

So that: abs[44 - 13] =31

Perfect will be a=b, because $\left |a-b \right |=> \left |a-a \right |=0$ so for $a=b$ we have:

$a^2 -6a + 5a - 373 =0$ <=> $a^2 -a - 373 =0$

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$=> $x = {1 \pm \sqrt{1-4(1)(-373)} \over 2}$=>$x = {1 \pm \sqrt{1493} \over 2}$=> $x = {1 \pm ~38.63 \over 2}$

$x1=\frac{39.63}{2} = 19.81$

$x2=\frac{-37.63}{2}=-18.81$ BUT because is negative rejected 

so $a=20$ so $20b-120+5b=373$ => b=19.72 its NOT integers so you try values until a,b  integers

Finaly the answer is a=44  and  b =13 like Guest says and can see it in this graph if you try values! 

https://www.desmos.com/calculator/4fhytbda6x

Hope it helps!

Thank y'all both! :)

ab - 6a + 5b  =  373

We can write this as

ab - 6a + 5b  - 30  =  373 - 30

(a + 5 )  ( b - 6 )  =  373 - 30

(a + 5) (b - 6)  =  343

343  as a product of two factors  = 

1  * 343    or  343 * 1       

7 * 49    or   49 * 7

So....we have the following possibilities for a, b

a           b

338      7

2         55

44       13

So....   l a - b l    minimized  =  l 44 - 13 l  =  31