H=-16t^2+285+2

Solution for H=-16t^2+285+2 equation:

=-16H^2+285+2

We move all terms to the left:

-(-16H^2+285+2)=0

We get rid of parentheses

16H^2-285-2=0

We add all the numbers together, and all the variables

16H^2-287=0

a = 16; b = 0; c = -287;
Δ = b2-4ac
Δ = 02-4·16·(-287)
Δ = 18368
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:

$\sqrt{\Delta}=\sqrt{18368}=\sqrt{64*287}=\sqrt{64}*\sqrt{287}=8\sqrt{287}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{287}}{2*16}=\frac{0-8\sqrt{287}}{32}
=-\frac{8\sqrt{287}}{32}
=-\frac{\sqrt{287}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{287}}{2*16}=\frac{0+8\sqrt{287}}{32}
=\frac{8\sqrt{287}}{32}
=\frac{\sqrt{287}}{4}
$