H=-4t^2+16t

Solution for H=-4t^2+16t equation:

=-4H^2+16H

We move all terms to the left:

-(-4H^2+16H)=0

We get rid of parentheses

4H^2-16H=0

a = 4; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·4·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*4}=\frac{0}{8}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*4}=\frac{32}{8}
=4
$