Ah, I didn't read your answer properly - that line should read
2n2+n−1≤n2 (letting C = 1), whence subtracting
2n2 from both sides will give you an inequality which holds for sufficiently large n.
What you have done *could* have been a proof - as long as you also show that equality holds with the inequality only for n<0. (You'd do that by solving the quadratic equation
2n2+n−1=n2.) That would then be like drawing a plot of the function (n against
2n(n+1−1−n2), and showing that it intersects the n-axis only for n<0; then observing that at n=0 the function lies above the n-axis, so it must also lie above the n-axis for n>0. That turns out to work in this case, because in fact the inequality is always satisfied for all real n, but it's a bit less safe as a method - it might not work so easily in all cases.