H=-16t^2+2t+1

Solution for H=-16t^2+2t+1 equation:

=-16H^2+2H+1

We move all terms to the left:

-(-16H^2+2H+1)=0

We get rid of parentheses

16H^2-2H-1=0

a = 16; b = -2; c = -1;
Δ = b2-4ac
Δ = -22-4·16·(-1)
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{17}}{2*16}=\frac{2-2\sqrt{17}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{17}}{2*16}=\frac{2+2\sqrt{17}}{32}
$