H=-16t2+28t+3

Solution for H=-16t2+28t+3 equation:

=-16H^2+28H+3

We move all terms to the left:

-(-16H^2+28H+3)=0

We get rid of parentheses

16H^2-28H-3=0

a = 16; b = -28; c = -3;
Δ = b2-4ac
Δ = -282-4·16·(-3)
Δ = 976
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:

$\sqrt{\Delta}=\sqrt{976}=\sqrt{16*61}=\sqrt{16}*\sqrt{61}=4\sqrt{61}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{61}}{2*16}=\frac{28-4\sqrt{61}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{61}}{2*16}=\frac{28+4\sqrt{61}}{32}
$