Y=-16t^2+4t+2

Solution for Y=-16t^2+4t+2 equation:

=-16Y^2+4Y+2

We move all terms to the left:

-(-16Y^2+4Y+2)=0

We get rid of parentheses

16Y^2-4Y-2=0

a = 16; b = -4; c = -2;
Δ = b2-4ac
Δ = -42-4·16·(-2)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*16}=\frac{-8}{32}
=-1/4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*16}=\frac{16}{32}
=1/2
$