H=-16t2+156t+105

Solution for H=-16t2+156t+105 equation:

=-16H^2+156H+105

We move all terms to the left:

-(-16H^2+156H+105)=0

We get rid of parentheses

16H^2-156H-105=0

a = 16; b = -156; c = -105;
Δ = b2-4ac
Δ = -1562-4·16·(-105)
Δ = 31056
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:

$\sqrt{\Delta}=\sqrt{31056}=\sqrt{16*1941}=\sqrt{16}*\sqrt{1941}=4\sqrt{1941}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-156)-4\sqrt{1941}}{2*16}=\frac{156-4\sqrt{1941}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-156)+4\sqrt{1941}}{2*16}=\frac{156+4\sqrt{1941}}{32}
$