H(t)=-18t^2+99t

Solution for H(t)=-18t^2+99t equation:

(H)=-18H^2+99H

We move all terms to the left:

(H)-(-18H^2+99H)=0

We get rid of parentheses

18H^2-99H+H=0

We add all the numbers together, and all the variables

18H^2-98H=0

a = 18; b = -98; c = 0;
Δ = b2-4ac
Δ = -982-4·18·0
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-98}{2*18}=\frac{0}{36}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+98}{2*18}=\frac{196}{36}
=5+4/9
$