H=-8t^2+120t-160

Solution for H=-8t^2+120t-160 equation:

=-8H^2+120H-160

We move all terms to the left:

-(-8H^2+120H-160)=0

We get rid of parentheses

8H^2-120H+160=0

a = 8; b = -120; c = +160;
Δ = b2-4ac
Δ = -1202-4·8·160
Δ = 9280
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:

$\sqrt{\Delta}=\sqrt{9280}=\sqrt{64*145}=\sqrt{64}*\sqrt{145}=8\sqrt{145}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-120)-8\sqrt{145}}{2*8}=\frac{120-8\sqrt{145}}{16}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-120)+8\sqrt{145}}{2*8}=\frac{120+8\sqrt{145}}{16}
$