H=-16t^2+t+50

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

=-16H^2+H+50

We move all terms to the left:

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

We get rid of parentheses

16H^2-H-50=0

We add all the numbers together, and all the variables

16H^2-1H-50=0

a = 16; b = -1; c = -50;
Δ = b2-4ac
Δ = -12-4·16·(-50)
Δ = 3201
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{-(-1)-\sqrt{3201}}{2*16}=\frac{1-\sqrt{3201}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{3201}}{2*16}=\frac{1+\sqrt{3201}}{32}
$