H(t)=-8t2+32t

Solution for H(t)=-8t2+32t equation:

(H)=-8H^2+32H

We move all terms to the left:

(H)-(-8H^2+32H)=0

We get rid of parentheses

8H^2-32H+H=0

We add all the numbers together, and all the variables

8H^2-31H=0

a = 8; b = -31; c = 0;
Δ = b2-4ac
Δ = -312-4·8·0
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-31}{2*8}=\frac{0}{16}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+31}{2*8}=\frac{62}{16}
=3+7/8
$