Y=-16t2+120

Solution for Y=-16t2+120 equation:

=-16Y^2+120

We move all terms to the left:

-(-16Y^2+120)=0

We get rid of parentheses

16Y^2-120=0

a = 16; b = 0; c = -120;
Δ = b2-4ac
Δ = 02-4·16·(-120)
Δ = 7680
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:

$\sqrt{\Delta}=\sqrt{7680}=\sqrt{256*30}=\sqrt{256}*\sqrt{30}=16\sqrt{30}$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{30}}{2*16}=\frac{0-16\sqrt{30}}{32}
=-\frac{16\sqrt{30}}{32}
=-\frac{\sqrt{30}}{2}
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{30}}{2*16}=\frac{0+16\sqrt{30}}{32}
=\frac{16\sqrt{30}}{32}
=\frac{\sqrt{30}}{2}
$