H=-16t^2+108t+160

Solution for H=-16t^2+108t+160 equation:

=-16H^2+108H+160

We move all terms to the left:

-(-16H^2+108H+160)=0

We get rid of parentheses

16H^2-108H-160=0

a = 16; b = -108; c = -160;
Δ = b2-4ac
Δ = -1082-4·16·(-160)
Δ = 21904
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{21904}=148$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-108)-148}{2*16}=\frac{-40}{32}
=-1+1/4
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-108)+148}{2*16}=\frac{256}{32}
=8
$