ahh, well to get that third solution is pretty sucky. Who gave you that question?
Also i really don't think you should approximate answers for the hsc. The only way i can see answering that is to use the lambert W function. Your answers are very close.
Lambert W function - Wikipedia, the free encyclopedia
2^x = 2x assuming x<0
1 = 2x/(2^x)
1 = 2x*e^(-x*ln2)
(-ln2)/2 = (-ln2*x)*e^(-x*ln2) this is in lambert form
-ln2*x = W(-ln2/2) W is the lambert W function
x = [-W(-ln2/2)]/ln2 this will give you your result
it might be worth noting that this uses the 3rd solution to x^2 = 2^x. Solutions 2, 4, -0.7666646959....
where x ln|2| = ln |-0.7666646959|
x = ln(0.7666646959)/ln2 = -0.383332348
there's probably a reason why but this isn't the place. I suspect there's a really easy solution using complex numbers. Sorry for babbling.