H(t)=-16t^2+120t+25

Solution for H(t)=-16t^2+120t+25 equation:

(H)=-16H^2+120H+25

We move all terms to the left:

(H)-(-16H^2+120H+25)=0

We get rid of parentheses

16H^2-120H+H-25=0

We add all the numbers together, and all the variables

16H^2-119H-25=0

a = 16; b = -119; c = -25;
Δ = b2-4ac
Δ = -1192-4·16·(-25)
Δ = 15761
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{-(-119)-\sqrt{15761}}{2*16}=\frac{119-\sqrt{15761}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-119)+\sqrt{15761}}{2*16}=\frac{119+\sqrt{15761}}{32}
$