Let $A$ be a subset of $mathbb{R}$ such that $A$ is bounded below with inf $A = L > 0$.

Since, $inf A = L > 0 $ , $ A $ is the set of positive reals, $(-a)x^2 + (-b)x > (-a)L^2 + (-b)L > 0 $
implies $ax^2 + bx < aL^2 + bL < 0 $
So, $aL^2 + bL $ is a upper bound for $B $. ( $B $ is the set of negative reals)
Now, we have to show that $aL^2 + bL $ is a least upper bound.
If possible let there is some upper bound $aM^2 + bM $ of $B $ such that $aM^2 + bM < aL^2 + bL < 0 $
this implies $(-a)M^2 + (-b)M > (-a)L^2 + (-b)L > 0 $
from these we can deduce that $ M > L $ and $ M $ is a lower bound of $ A $.(because if $ M $ is not a lower bound of $ A $, then $aM^2 + bM $ can't be a upper bound of $ B $).
So, clearly from $ M > L $ , we get a contradiction to the fact that $ L $ is greatest lower bound of $ A $.
So, our assumption is wrong.
Hence, $ aL^2 + bL = sup B$

No it's not correct. This inequality, $k > aL^2 + bL - epsilon$, cannot be derived for all $epsilon$. The respective inequality can be only derived for $sup B$, since that is the supreme.

You can assume thou that if $aL^2 + bL > sup B$ then by setting $epsilon_0= aL^2+bL-sup B>0$, you have that, there exist $x_0$ such that, $0>aL^2+bL geq ax_0^2+bx_0geq sup B-epsilon_0=-(aL^2+bL)> 0$, which is absurd