The equation v_{2}^{2} = v_{o}^{2} + 2aΔy is used to determine the change in position, acceleration, or velocity of a particle under the influence of a quadratic force. on earth, the acceleration value in this equation would, under normal circumstances, be a= -9.8 m/s^{2}. so, this substitutes into v_{2}^{2} = v_{o}^{2} + -19.6Δy In this Δy = v_{2}^{2} - v_{o}^{2} / -19.6 Lets now assume that v_{2}^{2} is 0, the crest of a projectiles motion. Δy = 0 - v_{o}^{2} / -19.6 If the objects initial v is facing upwards, then obviously the net Δy would be positive. Now, keeping all conditions the same, lets assume that the acceleration is 9.8m/s^{s} upwards This would yield Δy = 0 - v_{o}^{2} / 19.6 Δy would thus be negative, and the object would be thrown downwards, accelerating upwards.
I just want to go through your reasoning to make sure I follow: That would be a net unbalanced force which is constant. "normal circumstances" being, +y = "upwards", Δy<<R_{Earth}, and gravity is the only force acting on the object. Your value has changed somehow [x2 of course] ... better to put g=9.8N/kg so a=-g so it does not matter what units you want to work in just yet... you get:$$v_2^2=v_0^2-2g\Delta y$$ $$\Delta y = \frac{v_2^2-v_0^2}{-2g} = \frac{v_0^2-v_2^2}{2g}$$ $$\Delta y = \frac{v_0^2}{2g} > 0$$ No problems there. All conditions? So you put gravity upwards, and the final velocity is zero, but the assumption that the initial velocity is also upwards is incorrect - as you found below: $$\Delta y = \frac{0-v_0^2}{2g}=-\frac{v_0^2}{2g} < 0$$ Well ... you did specify that the final velocity is zero ... if the object were thrown upwards with an upwards acceleration, that would not be the case would it?
When an object is thrown upwards, the acceleration is still 9.8 m/s^{2} downwards, but there is now an initial upward velocity imposed on the object which was simply not present in the case where the object was initially at rest and allowed to drop.