H=-16t2+2704

Solution for H=-16t2+2704 equation:

=-16H^2+2704

We move all terms to the left:

-(-16H^2+2704)=0

We get rid of parentheses

16H^2-2704=0

a = 16; b = 0; c = -2704;
Δ = b2-4ac
Δ = 02-4·16·(-2704)
Δ = 173056
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{173056}=416$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-416}{2*16}=\frac{-416}{32}
=-13
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+416}{2*16}=\frac{416}{32}
=13
$