H=-16t2+34t+100

Solution for H=-16t2+34t+100 equation:

=-16H^2+34H+100

We move all terms to the left:

-(-16H^2+34H+100)=0

We get rid of parentheses

16H^2-34H-100=0

a = 16; b = -34; c = -100;
Δ = b2-4ac
Δ = -342-4·16·(-100)
Δ = 7556
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:

$\sqrt{\Delta}=\sqrt{7556}=\sqrt{4*1889}=\sqrt{4}*\sqrt{1889}=2\sqrt{1889}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-34)-2\sqrt{1889}}{2*16}=\frac{34-2\sqrt{1889}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-34)+2\sqrt{1889}}{2*16}=\frac{34+2\sqrt{1889}}{32}
$