H=2+28t-16t^2

Solution for H=2+28t-16t^2 equation:

=2+28H-16H^2

We move all terms to the left:

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

We get rid of parentheses

16H^2-28H-2=0

a = 16; b = -28; c = -2;
Δ = b2-4ac
Δ = -282-4·16·(-2)
Δ = 912
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:

$\sqrt{\Delta}=\sqrt{912}=\sqrt{16*57}=\sqrt{16}*\sqrt{57}=4\sqrt{57}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{57}}{2*16}=\frac{28-4\sqrt{57}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{57}}{2*16}=\frac{28+4\sqrt{57}}{32}
$