Y=-16t^2+160t+4

Solution for Y=-16t^2+160t+4 equation:

=-16Y^2+160Y+4

We move all terms to the left:

-(-16Y^2+160Y+4)=0

We get rid of parentheses

16Y^2-160Y-4=0

a = 16; b = -160; c = -4;
Δ = b2-4ac
Δ = -1602-4·16·(-4)
Δ = 25856
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:

$\sqrt{\Delta}=\sqrt{25856}=\sqrt{256*101}=\sqrt{256}*\sqrt{101}=16\sqrt{101}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-16\sqrt{101}}{2*16}=\frac{160-16\sqrt{101}}{32}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+16\sqrt{101}}{2*16}=\frac{160+16\sqrt{101}}{32}
$