Resistor values for OpAmp circuit with dc offset applied to inverting pin
First of all your transfer function is wrong. 

From what I see the correct one should look like this:

$$V_{OUT} = V_{IN}\left(1 + \frac{R_F}{R_1||R_2}\right) - V_{REF}\left(\frac{R_F}{R_1}\right) $$ 

For this circuit 



Now let us assume that you want "shift" the voltage from \$0V ... 3.3V\$ into \$-40V...+40V\$

Therefore the "inverting part" of a voltage gain is \$ \frac{- V_{OUT}}{V_{REF}}=  -\frac{40V}{3.3V} = -12.121V/V\$

So I pick \$R_1 = 75 \textrm{k}\Omega , R_F = 910\textrm{k}\Omega\$

http://jansson.us/resistors.html

http://www.ti.com/download/kbase/volt/volt_div3.htm

Now we can solve for \$R_2\$ by using this case:



Notice that \$R_1\$ current is \$0A\$ hence solving for \$R_2\$ resistor value is a trivial task. 

$$R_2 = \frac{R_F}{\frac{V_{OUT}}{V_{IN}} -1} = \frac{910\textrm{k}\Omega}{\frac{40V}{3.3V} -1} = 82\textrm{k}\Omega $$

And we done.

Additional bedtime reading:

http://www.ti.com/lit/an/sloa030a/sloa030a.pdf