H(t)=-16t^2+440

Solution for H(t)=-16t^2+440 equation:

(H)=-16H^2+440

We move all terms to the left:

(H)-(-16H^2+440)=0

We get rid of parentheses

16H^2+H-440=0

a = 16; b = 1; c = -440;
Δ = b2-4ac
Δ = 12-4·16·(-440)
Δ = 28161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28161}=\sqrt{9*3129}=\sqrt{9}*\sqrt{3129}=3\sqrt{3129}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{3129}}{2*16}=\frac{-1-3\sqrt{3129}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{3129}}{2*16}=\frac{-1+3\sqrt{3129}}{32}
$