H=-16t^2+160t+3

Solution for H=-16t^2+160t+3 equation:

=-16H^2+160H+3

We move all terms to the left:

-(-16H^2+160H+3)=0

We get rid of parentheses

16H^2-160H-3=0

a = 16; b = -160; c = -3;
Δ = b2-4ac
Δ = -1602-4·16·(-3)
Δ = 25792
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:

$\sqrt{\Delta}=\sqrt{25792}=\sqrt{64*403}=\sqrt{64}*\sqrt{403}=8\sqrt{403}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{403}}{2*16}=\frac{160-8\sqrt{403}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{403}}{2*16}=\frac{160+8\sqrt{403}}{32}
$