H=16t2+32t+6

Solution for H=16t2+32t+6 equation:

=16H^2+32H+6

We move all terms to the left:

-(16H^2+32H+6)=0

We get rid of parentheses

-16H^2-32H-6=0

a = -16; b = -32; c = -6;
Δ = b2-4ac
Δ = -322-4·(-16)·(-6)
Δ = 640
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:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{10}}{2*-16}=\frac{32-8\sqrt{10}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{10}}{2*-16}=\frac{32+8\sqrt{10}}{-32}
$