(H)=-162t^2+32t+48

Solution for (H)=-162t^2+32t+48 equation:

()=-162H^2+32H+48

We move all terms to the left:

()-(-162H^2+32H+48)=0

We add all the numbers together, and all the variables

-(-162H^2+32H+48)=0

We get rid of parentheses

162H^2-32H-48=0

a = 162; b = -32; c = -48;
Δ = b2-4ac
Δ = -322-4·162·(-48)
Δ = 32128
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:

$\sqrt{\Delta}=\sqrt{32128}=\sqrt{64*502}=\sqrt{64}*\sqrt{502}=8\sqrt{502}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{502}}{2*162}=\frac{32-8\sqrt{502}}{324}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{502}}{2*162}=\frac{32+8\sqrt{502}}{324}
$