Y=-16t^2+32t+3

Solution for Y=-16t^2+32t+3 equation:

=-16Y^2+32Y+3

We move all terms to the left:

-(-16Y^2+32Y+3)=0

We get rid of parentheses

16Y^2-32Y-3=0

a = 16; b = -32; c = -3;
Δ = b2-4ac
Δ = -322-4·16·(-3)
Δ = 1216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{19}}{2*16}=\frac{32-8\sqrt{19}}{32}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{19}}{2*16}=\frac{32+8\sqrt{19}}{32}
$