H(t)=-16t^2+320t+45

Solution for H(t)=-16t^2+320t+45 equation:

(H)=-16H^2+320H+45

We move all terms to the left:

(H)-(-16H^2+320H+45)=0

We get rid of parentheses

16H^2-320H+H-45=0

We add all the numbers together, and all the variables

16H^2-319H-45=0

a = 16; b = -319; c = -45;
Δ = b2-4ac
Δ = -3192-4·16·(-45)
Δ = 104641
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{104641}}{2*16}=\frac{319-\sqrt{104641}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-319)+\sqrt{104641}}{2*16}=\frac{319+\sqrt{104641}}{32}
$