H=-16t^2+127t+50

Solution for H=-16t^2+127t+50 equation:

=-16H^2+127H+50

We move all terms to the left:

-(-16H^2+127H+50)=0

We get rid of parentheses

16H^2-127H-50=0

a = 16; b = -127; c = -50;
Δ = b2-4ac
Δ = -1272-4·16·(-50)
Δ = 19329
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{-(-127)-\sqrt{19329}}{2*16}=\frac{127-\sqrt{19329}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-127)+\sqrt{19329}}{2*16}=\frac{127+\sqrt{19329}}{32}
$