H=-16t^2+240

Solution for H=-16t^2+240 equation:

=-16H^2+240

We move all terms to the left:

-(-16H^2+240)=0

We get rid of parentheses

16H^2-240=0

a = 16; b = 0; c = -240;
Δ = b2-4ac
Δ = 02-4·16·(-240)
Δ = 15360
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{15}}{2*16}=\frac{0-32\sqrt{15}}{32}
=-\frac{32\sqrt{15}}{32}
=-\sqrt{15}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{15}}{2*16}=\frac{0+32\sqrt{15}}{32}
=\frac{32\sqrt{15}}{32}
=\sqrt{15}
$