How do you arrive at that ? I thought i had to find the equivalent voltage by looking "into" the load terminals where i would see the two voltage sources as one as I "probe" with an infinite impedence and then the source impedence is the resulting impedence having shorted the voltage sources which puts the two inductors in parallel.

I'm sorry if I'm being opaque, I was writing from a tablet. When computing the equivalent voltage, the voltage sources remain (as opposed to when computing the resistance), so the inductors are not exactly in parallel, because each one is connected to a different voltage, v1 and v2. The best way of seeing it is that your infinite impedance probe is at the middle of a voltage divider between v1 and v2.

To complete the computation:

Veq = 166 + 249J

Req = 2.4 Ohm

The resistor sees a voltage v_load = Veq * Zload / (Zload + Req) = 152.6 + 245.9J (Zload = 35.5 + 35.7J)

The current through the load is I = v_load / Zload = 5.6 + 1.3J, with modulus |I| = 5.75 A.

I think that's closer to the simulation.

Edit: in other words, you see the two voltage sources, but each loads the other, so from the middle point you see a middle voltage. The voltages don't simply add, nor they average, because the impedances looking into each one are different.