H(t)=-16t^2+20t+8

Solution for H(t)=-16t^2+20t+8 equation:

(H)=-16H^2+20H+8

We move all terms to the left:

(H)-(-16H^2+20H+8)=0

We get rid of parentheses

16H^2-20H+H-8=0

We add all the numbers together, and all the variables

16H^2-19H-8=0

a = 16; b = -19; c = -8;
Δ = b2-4ac
Δ = -192-4·16·(-8)
Δ = 873
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:

$\sqrt{\Delta}=\sqrt{873}=\sqrt{9*97}=\sqrt{9}*\sqrt{97}=3\sqrt{97}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-3\sqrt{97}}{2*16}=\frac{19-3\sqrt{97}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+3\sqrt{97}}{2*16}=\frac{19+3\sqrt{97}}{32}
$