H(t)=-16t^2+480t+90

Solution for H(t)=-16t^2+480t+90 equation:

(H)=-16H^2+480H+90

We move all terms to the left:

(H)-(-16H^2+480H+90)=0

We get rid of parentheses

16H^2-480H+H-90=0

We add all the numbers together, and all the variables

16H^2-479H-90=0

a = 16; b = -479; c = -90;
Δ = b2-4ac
Δ = -4792-4·16·(-90)
Δ = 235201
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{-(-479)-\sqrt{235201}}{2*16}=\frac{479-\sqrt{235201}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-479)+\sqrt{235201}}{2*16}=\frac{479+\sqrt{235201}}{32}
$