H=-16t^2+105t+8

Solution for H=-16t^2+105t+8 equation:

=-16H^2+105H+8

We move all terms to the left:

-(-16H^2+105H+8)=0

We get rid of parentheses

16H^2-105H-8=0

a = 16; b = -105; c = -8;
Δ = b2-4ac
Δ = -1052-4·16·(-8)
Δ = 11537
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-105)-\sqrt{11537}}{2*16}=\frac{105-\sqrt{11537}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-105)+\sqrt{11537}}{2*16}=\frac{105+\sqrt{11537}}{32}
$