H=-16t^2+34t+528

Solution for H=-16t^2+34t+528 equation:

=-16H^2+34H+528

We move all terms to the left:

-(-16H^2+34H+528)=0

We get rid of parentheses

16H^2-34H-528=0

a = 16; b = -34; c = -528;
Δ = b2-4ac
Δ = -342-4·16·(-528)
Δ = 34948
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:

$\sqrt{\Delta}=\sqrt{34948}=\sqrt{4*8737}=\sqrt{4}*\sqrt{8737}=2\sqrt{8737}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2\sqrt{8737}}{2*16}=\frac{34-2\sqrt{8737}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2\sqrt{8737}}{2*16}=\frac{34+2\sqrt{8737}}{32}
$