H=-16t2+50

Solution for H=-16t2+50 equation:

=-16H^2+50

We move all terms to the left:

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

We get rid of parentheses

16H^2-50=0

a = 16; b = 0; c = -50;
Δ = b2-4ac
Δ = 02-4·16·(-50)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*16}=\frac{0-40\sqrt{2}}{32}
=-\frac{40\sqrt{2}}{32}
=-\frac{5\sqrt{2}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*16}=\frac{0+40\sqrt{2}}{32}
=\frac{40\sqrt{2}}{32}
=\frac{5\sqrt{2}}{4}
$