H=80t+16t^2

Solution for H=80t+16t^2 equation:

=80H+16H^2

We move all terms to the left:

-(80H+16H^2)=0

We get rid of parentheses

-16H^2-80H=0

a = -16; b = -80; c = 0;
Δ = b2-4ac
Δ = -802-4·(-16)·0
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-80}{2*-16}=\frac{0}{-32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+80}{2*-16}=\frac{160}{-32}
=-5
$