H(t)=16t^2+54t+7

Solution for H(t)=16t^2+54t+7 equation:

(H)=16H^2+54H+7

We move all terms to the left:

(H)-(16H^2+54H+7)=0

We get rid of parentheses

-16H^2+H-54H-7=0

We add all the numbers together, and all the variables

-16H^2-53H-7=0

a = -16; b = -53; c = -7;
Δ = b2-4ac
Δ = -532-4·(-16)·(-7)
Δ = 2361
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{2361}}{2*-16}=\frac{53-\sqrt{2361}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-53)+\sqrt{2361}}{2*-16}=\frac{53+\sqrt{2361}}{-32}
$