H=-16t^2+102t+14

Solution for H=-16t^2+102t+14 equation:

=-16H^2+102H+14

We move all terms to the left:

-(-16H^2+102H+14)=0

We get rid of parentheses

16H^2-102H-14=0

a = 16; b = -102; c = -14;
Δ = b2-4ac
Δ = -1022-4·16·(-14)
Δ = 11300
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:

$\sqrt{\Delta}=\sqrt{11300}=\sqrt{100*113}=\sqrt{100}*\sqrt{113}=10\sqrt{113}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-102)-10\sqrt{113}}{2*16}=\frac{102-10\sqrt{113}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-102)+10\sqrt{113}}{2*16}=\frac{102+10\sqrt{113}}{32}
$