H=-16t^2+122t+5

Solution for H=-16t^2+122t+5 equation:

=-16H^2+122H+5

We move all terms to the left:

-(-16H^2+122H+5)=0

We get rid of parentheses

16H^2-122H-5=0

a = 16; b = -122; c = -5;
Δ = b2-4ac
Δ = -1222-4·16·(-5)
Δ = 15204
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:

$\sqrt{\Delta}=\sqrt{15204}=\sqrt{4*3801}=\sqrt{4}*\sqrt{3801}=2\sqrt{3801}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-122)-2\sqrt{3801}}{2*16}=\frac{122-2\sqrt{3801}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-122)+2\sqrt{3801}}{2*16}=\frac{122+2\sqrt{3801}}{32}
$