H(t)=-16t^2+54t+30

Solution for H(t)=-16t^2+54t+30 equation:

(H)=-16H^2+54H+30

We move all terms to the left:

(H)-(-16H^2+54H+30)=0

We get rid of parentheses

16H^2-54H+H-30=0

We add all the numbers together, and all the variables

16H^2-53H-30=0

a = 16; b = -53; c = -30;
Δ = b2-4ac
Δ = -532-4·16·(-30)
Δ = 4729
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{-(-53)-\sqrt{4729}}{2*16}=\frac{53-\sqrt{4729}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+\sqrt{4729}}{2*16}=\frac{53+\sqrt{4729}}{32}
$