H(t)=-16t^2+3.5t

Solution for H(t)=-16t^2+3.5t equation:

(H)=-16H^2+3.5H

We move all terms to the left:

(H)-(-16H^2+3.5H)=0

We get rid of parentheses

16H^2-3.5H+H=0

We add all the numbers together, and all the variables

16H^2-2.5H=0

a = 16; b = -2.5; c = 0;
Δ = b2-4ac
Δ = -2.52-4·16·0
Δ = 6.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2.5)-\sqrt{6.25}}{2*16}=\frac{2.5-\sqrt{6.25}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2.5)+\sqrt{6.25}}{2*16}=\frac{2.5+\sqrt{6.25}}{32}
$