Question

How do you factor and solve `6b^2 - 13b + 3 = -3`?

Answer 1

Rewrite `6b^2 - 13b + 3 = -3`
as
`6b^2-13b+6=0`

Assuming only integer values appear in the factoring (not necessarily true, but easier if it is)
we are looking for integer values for `p`, `q`, `r`, and `s` (all `>= 0`)
such that
`(pb-q)(rb-s) = 6b^2-13b+6`
(the negative sign on the coefficient `-13` tells us that at least one of the internal signs in the factors is negative, assuming positive values for `p, q, r, s`
and the positive sign on `+6` tells us that the internal signs are the same i.e. they are both negative).

`(pb-q)(rb-s) = (pbr)b^2 - (ps+qr)b + qs`
`rarr pb = 6`
`rarr ps+qr = 13`
`rarr qs = 6`

Since the only positive integer factors of `6` are `6xx1` and `3xx2`

It is a fairly simply matter to discover
`p=3`
`q=2`
`r=3`
`s=2`

That is
`(3b-2)^2 = 6b^2-12b+6 = 0`

Which implies there is only one solution (`3b-2=0`)
`b=2/3`