H(t)=-16(t)^2+144

Solution for H(t)=-16(t)^2+144 equation:

(H)=-16(H)^2+144

We move all terms to the left:

(H)-(-16(H)^2+144)=0

determiningTheFunctionDomain
-(-16H^2+144)+H=0

We get rid of parentheses

16H^2+H-144=0

a = 16; b = 1; c = -144;
Δ = b2-4ac
Δ = 12-4·16·(-144)
Δ = 9217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{9217}}{2*16}=\frac{-1-\sqrt{9217}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{9217}}{2*16}=\frac{-1+\sqrt{9217}}{32}
$