H(t)=-16t^2+104t+56.

Solution for H(t)=-16t^2+104t+56. equation:

(H)=-16H^2+104H+56.

We move all terms to the left:

(H)-(-16H^2+104H+56.)=0

We get rid of parentheses

16H^2-104H+H-56.=0

We add all the numbers together, and all the variables

16H^2-103H-56=0

a = 16; b = -103; c = -56;
Δ = b2-4ac
Δ = -1032-4·16·(-56)
Δ = 14193
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:

$\sqrt{\Delta}=\sqrt{14193}=\sqrt{9*1577}=\sqrt{9}*\sqrt{1577}=3\sqrt{1577}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-103)-3\sqrt{1577}}{2*16}=\frac{103-3\sqrt{1577}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-103)+3\sqrt{1577}}{2*16}=\frac{103+3\sqrt{1577}}{32}
$