H(t)=-16t^2+320t+5

Solution for H(t)=-16t^2+320t+5 equation:

(H)=-16H^2+320H+5

We move all terms to the left:

(H)-(-16H^2+320H+5)=0

We get rid of parentheses

16H^2-320H+H-5=0

We add all the numbers together, and all the variables

16H^2-319H-5=0

a = 16; b = -319; c = -5;
Δ = b2-4ac
Δ = -3192-4·16·(-5)
Δ = 102081
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{-(-319)-\sqrt{102081}}{2*16}=\frac{319-\sqrt{102081}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-319)+\sqrt{102081}}{2*16}=\frac{319+\sqrt{102081}}{32}
$