Y=-16t^2+408

Solution for Y=-16t^2+408 equation:

=-16Y^2+408

We move all terms to the left:

-(-16Y^2+408)=0

We get rid of parentheses

16Y^2-408=0

a = 16; b = 0; c = -408;
Δ = b2-4ac
Δ = 02-4·16·(-408)
Δ = 26112
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:

$\sqrt{\Delta}=\sqrt{26112}=\sqrt{256*102}=\sqrt{256}*\sqrt{102}=16\sqrt{102}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{102}}{2*16}=\frac{0-16\sqrt{102}}{32}
=-\frac{16\sqrt{102}}{32}
=-\frac{\sqrt{102}}{2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{102}}{2*16}=\frac{0+16\sqrt{102}}{32}
=\frac{16\sqrt{102}}{32}
=\frac{\sqrt{102}}{2}
$