H(t)=-16t2+18t+120

Solution for H(t)=-16t2+18t+120 equation:

(H)=-16H^2+18H+120

We move all terms to the left:

(H)-(-16H^2+18H+120)=0

We get rid of parentheses

16H^2-18H+H-120=0

We add all the numbers together, and all the variables

16H^2-17H-120=0

a = 16; b = -17; c = -120;
Δ = b2-4ac
Δ = -172-4·16·(-120)
Δ = 7969
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{-(-17)-\sqrt{7969}}{2*16}=\frac{17-\sqrt{7969}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{7969}}{2*16}=\frac{17+\sqrt{7969}}{32}
$