H=-16t2+62t+2

Solution for H=-16t2+62t+2 equation:

=-16H^2+62H+2

We move all terms to the left:

-(-16H^2+62H+2)=0

We get rid of parentheses

16H^2-62H-2=0

a = 16; b = -62; c = -2;
Δ = b2-4ac
Δ = -622-4·16·(-2)
Δ = 3972
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:

$\sqrt{\Delta}=\sqrt{3972}=\sqrt{4*993}=\sqrt{4}*\sqrt{993}=2\sqrt{993}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-2\sqrt{993}}{2*16}=\frac{62-2\sqrt{993}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+2\sqrt{993}}{2*16}=\frac{62+2\sqrt{993}}{32}
$