Strain Gauge with Instrumentation Amplifier question

I have a problem making a circuit like this working

I got it from this TI application note. For testing purposes I have set:

$R4=R2=R3=R1=47 \Omega$

$ Rf = 10 k\Omega$

$Rg= 26.70\Omega$

The op amp $V_S=+5V$. This should give a gain of 750 having $V_{SIG}$ as 0.8E-4V/kg.

I agreed that 80kg would be enough for full scale. The problem is that this setup doesn't show any variations when i load the strain gauge. It just stays at 2.40V which doesn't make much sense since it doesn't reach near $V_S$. I am using the TL064. I have tried a lot and think my wiring is correct. Do you have any suggestions?

Update: I came up with the 750 of gain by assuming that:

$R4=R2 <=> \dfrac{R4}{R2}=1$

$Gain=\dfrac{2Rf}{Rg}+1$
$(Sig_+)-(Sig_-) = V_{IN} => V_O=V_{IN} \times 1 \times Gain <=>$
$Gain = \dfrac{V_O}{V_{IN}} <=> \dfrac{2Rf}{Rg}+1 = \dfrac{V_O}{V_{IN}}.$
$V_{IN}/kg = 0.8E-4V/kg$

I also i assumed that my absolute maximum load would be 80kg. So $V_{IN(MAX)} = 0.8E-4 \times 80 = 6.4E-3$ V

So we can determine the needed gain for

$V_{OUT}=5V. G=\dfrac{V_{OUT}}{V_{IN}}=\dfrac{4.8}{6.4E-3}=750$

Taking the previous equations: $ \dfrac{2Rf}{Rg}+1 = 750 <=> 2Rf = 749 Rg <=> Rg = \dfrac{Rf}{374.5}$

Knowing this and the resistors I had available I chose to set Rg as a function of Rf because i have to arrange the resistors(series or parallels to approximately get the desired gain) I chose:

Edit:

Corrected Equation mistake but still no results $Rf=10 k\Omega => Rg = 10E3/374.5 = 26.70 \Omega$ Considering I don't have this precise value I came up with $Rg = 47 || 47 = 23.5$ which is close to $26.70 \Omega$

Hope everything is clear for everybody, otherwise please ask for clarifications Thank you