Electric Current: Resistance and Power?

Ok, P=IV, thus P=(V^2)/R. But every time I use that for the following problem I wind up getting the wrong answer.

The question reads: “A power station delivers 620 kW of power at 12,000 V to a factory through wires with 3 Ohms of total resistance. How much less power is wasted if the electricity is delivered at 50,000 V?

Provided answer is 7500 W but I’m getting negative answers in the 700 kW area… how do I set this problem up?

Power to be delivered P = V I = 620000 w
1) at V =12000 volts
current drawn I = P/V = 51.67 ampere
power loss p1 = R I^2 = = 3*(51.67)^2 = 8009.3 w
—————————–
2) at V =50000 volts
current drawn I = P/V = 620000/50000 = 12.4 ampere
power loss p2 = R I^2 = = 3*(12.4)^2 = 462.3 w
——————-
power wastage SAVED = p1 – p2 = 8009.3 – 462.3
= 7567 watt

Hendrik Stiesdal answers Wind Power Questions (Part 4)