H=-16t^2+160t+10

Solution for H=-16t^2+160t+10 equation:

=-16H^2+160H+10

We move all terms to the left:

-(-16H^2+160H+10)=0

We get rid of parentheses

16H^2-160H-10=0

a = 16; b = -160; c = -10;
Δ = b2-4ac
Δ = -1602-4·16·(-10)
Δ = 26240
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:

$\sqrt{\Delta}=\sqrt{26240}=\sqrt{64*410}=\sqrt{64}*\sqrt{410}=8\sqrt{410}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{410}}{2*16}=\frac{160-8\sqrt{410}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{410}}{2*16}=\frac{160+8\sqrt{410}}{32}
$