H(t)=-16t^2+20t+7

Solution for H(t)=-16t^2+20t+7 equation:

(H)=-16H^2+20H+7

We move all terms to the left:

(H)-(-16H^2+20H+7)=0

We get rid of parentheses

16H^2-20H+H-7=0

We add all the numbers together, and all the variables

16H^2-19H-7=0

a = 16; b = -19; c = -7;
Δ = b2-4ac
Δ = -192-4·16·(-7)
Δ = 809
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{-(-19)-\sqrt{809}}{2*16}=\frac{19-\sqrt{809}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{809}}{2*16}=\frac{19+\sqrt{809}}{32}
$