H(t)=16t^2+32t+4

Solution for H(t)=16t^2+32t+4 equation:

(H)=16H^2+32H+4

We move all terms to the left:

(H)-(16H^2+32H+4)=0

We get rid of parentheses

-16H^2+H-32H-4=0

We add all the numbers together, and all the variables

-16H^2-31H-4=0

a = -16; b = -31; c = -4;
Δ = b2-4ac
Δ = -312-4·(-16)·(-4)
Δ = 705
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{-(-31)-\sqrt{705}}{2*-16}=\frac{31-\sqrt{705}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{705}}{2*-16}=\frac{31+\sqrt{705}}{-32}
$