H(t)=-16(t^2)+16t+480

Solution for H(t)=-16(t^2)+16t+480 equation:

(H)=-16(H^2)+16H+480

We move all terms to the left:

(H)-(-16(H^2)+16H+480)=0

determiningTheFunctionDomain
-(-16H^2+16H+480)+H=0

We get rid of parentheses

16H^2-16H+H-480=0

We add all the numbers together, and all the variables

16H^2-15H-480=0

a = 16; b = -15; c = -480;
Δ = b2-4ac
Δ = -152-4·16·(-480)
Δ = 30945
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{30945}}{2*16}=\frac{15-\sqrt{30945}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{30945}}{2*16}=\frac{15+\sqrt{30945}}{32}
$