H(t)=-16t^2+t+5

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

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

We move all terms to the left:

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

We get rid of parentheses

16H^2-H+H-5=0

We add all the numbers together, and all the variables

16H^2-5=0

a = 16; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·16·(-5)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{5}}{2*16}=\frac{0-8\sqrt{5}}{32}
=-\frac{8\sqrt{5}}{32}
=-\frac{\sqrt{5}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{5}}{2*16}=\frac{0+8\sqrt{5}}{32}
=\frac{8\sqrt{5}}{32}
=\frac{\sqrt{5}}{4}
$