H=4t2+24t+28

Solution for H=4t2+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 = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{2}}{2*-4}=\frac{24-8\sqrt{2}}{-8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{2}}{2*-4}=\frac{24+8\sqrt{2}}{-8}
$