Y=-16t^2+208

Solution for Y=-16t^2+208 equation:

=-16Y^2+208

We move all terms to the left:

-(-16Y^2+208)=0

We get rid of parentheses

16Y^2-208=0

a = 16; b = 0; c = -208;
Δ = b2-4ac
Δ = 02-4·16·(-208)
Δ = 13312
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:

$\sqrt{\Delta}=\sqrt{13312}=\sqrt{1024*13}=\sqrt{1024}*\sqrt{13}=32\sqrt{13}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{13}}{2*16}=\frac{0-32\sqrt{13}}{32}
=-\frac{32\sqrt{13}}{32}
=-\sqrt{13}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{13}}{2*16}=\frac{0+32\sqrt{13}}{32}
=\frac{32\sqrt{13}}{32}
=\sqrt{13}
$