H=-4t^2+24t+28

Solution for H=-4t^2+24t+28 equation:

=-4H^2+24H+28

We move all terms to the left:

-(-4H^2+24H+28)=0

We get rid of parentheses

4H^2-24H-28=0

a = 4; b = -24; c = -28;
Δ = b2-4ac
Δ = -242-4·4·(-28)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-32}{2*4}=\frac{-8}{8}
=-1
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+32}{2*4}=\frac{56}{8}
=7
$