The solutions are: #-1/3#, #-9#

#3w^2+28w+9=0#

If it can be factored using whole numbers it must be

#(3w+"something"_1)(w+"something"_2)#

So that the #w^2# term has a coefficient of #3#

#"something"_1 xx "something"_2# must be #9#

(the product of the constants (the lasts) must be the constant #9#)

To get a product of #9# using whole numbers our choices are

#1*9# or #3*3#.

But the first terms are different so we need to consider the choices:

#(3w+1)(w+9)#

#(3w+9)(w+1)#

#(3w+3)(w+3)# (Notice this would make the whole thing divisible by #3# -- which it isn't. This won't work.)

A quick check will convince you that

#(3w+1)(w+9)=3w^2+28w+9#

So, solving #3w^2+28w+9=0# is the same as solving:

#(3w+1)(w+9)=0#

And the only way for a product of numbers to be #0# is for at least one of the numbers to be #0#. So we need:

#3w+1=0# or #w+9=0#

In the first case we get #w=-1/3#, in the second #w=-9#.

The solutions are: #-1/3#, #-9#