H(t)=-16t^2+16t+483

Solution for H(t)=-16t^2+16t+483 equation:

(H)=-16H^2+16H+483

We move all terms to the left:

(H)-(-16H^2+16H+483)=0

We get rid of parentheses

16H^2-16H+H-483=0

We add all the numbers together, and all the variables

16H^2-15H-483=0

a = 16; b = -15; c = -483;
Δ = b2-4ac
Δ = -152-4·16·(-483)
Δ = 31137
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{-(-15)-\sqrt{31137}}{2*16}=\frac{15-\sqrt{31137}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{31137}}{2*16}=\frac{15+\sqrt{31137}}{32}
$