H(t)=-16t^2+50+3

Solution for H(t)=-16t^2+50+3 equation:

(H)=-16H^2+50+3

We move all terms to the left:

(H)-(-16H^2+50+3)=0

We get rid of parentheses

16H^2+H-50-3=0

We add all the numbers together, and all the variables

16H^2+H-53=0

a = 16; b = 1; c = -53;
Δ = b2-4ac
Δ = 12-4·16·(-53)
Δ = 3393
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:

$\sqrt{\Delta}=\sqrt{3393}=\sqrt{9*377}=\sqrt{9}*\sqrt{377}=3\sqrt{377}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{377}}{2*16}=\frac{-1-3\sqrt{377}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{377}}{2*16}=\frac{-1+3\sqrt{377}}{32}
$