H=-16t2+6t+28

Solution for H=-16t2+6t+28 equation:

=-16H^2+6H+28

We move all terms to the left:

-(-16H^2+6H+28)=0

We get rid of parentheses

16H^2-6H-28=0

a = 16; b = -6; c = -28;
Δ = b2-4ac
Δ = -62-4·16·(-28)
Δ = 1828
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:

$\sqrt{\Delta}=\sqrt{1828}=\sqrt{4*457}=\sqrt{4}*\sqrt{457}=2\sqrt{457}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{457}}{2*16}=\frac{6-2\sqrt{457}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{457}}{2*16}=\frac{6+2\sqrt{457}}{32}
$