H(3)=-16t^2+56t+3

Solution for H(3)=-16t^2+56t+3 equation:

(3)=-16H^2+56H+3

We move all terms to the left:

(3)-(-16H^2+56H+3)=0

We get rid of parentheses

16H^2-56H-3+3=0

We add all the numbers together, and all the variables

16H^2-56H=0

a = 16; b = -56; c = 0;
Δ = b2-4ac
Δ = -562-4·16·0
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-56}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+56}{2*16}=\frac{112}{32}
=3+1/2
$