H(t)=-16(t)^2+100(t)+8

Solution for H(t)=-16(t)^2+100(t)+8 equation:

(H)=-16(H)^2+100(H)+8

We move all terms to the left:

(H)-(-16(H)^2+100(H)+8)=0

determiningTheFunctionDomain
-(-16H^2+100H+8)+H=0

We get rid of parentheses

16H^2-100H+H-8=0

We add all the numbers together, and all the variables

16H^2-99H-8=0

a = 16; b = -99; c = -8;
Δ = b2-4ac
Δ = -992-4·16·(-8)
Δ = 10313
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{-(-99)-\sqrt{10313}}{2*16}=\frac{99-\sqrt{10313}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-99)+\sqrt{10313}}{2*16}=\frac{99+\sqrt{10313}}{32}
$