Make a two-resistor voltage divider from a fixed voltage (supply) to ground, that produces 0.9volt.
Calculate the resistance of those two resisors in parallel.
Add a third resistor from the +10/-10 source to the tap that drops 10volt to 0.9volt.
Leo..
That's not quite correct because you want 0.9V when the input is 0V, not when the input is open circuit.
If the resistor to ground is 1kΩ, then the resistor to a 5V supply should be 4.056kΩ and the resistor to the input should be 8.093kΩ (I hope those figures are correct!). The values can be scaled, for example multiplying all three values by ten. Of course this assumes the 5V supply is accurate and very stable; also that the input to the analogue-to-digital converter is high impedance. If a high input impedance is required at the input of the resistor network, an operational amplifier unity-gain buffer could be used.