H(t)=-18t^2+124t

Solution for H(t)=-18t^2+124t equation:

(H)=-18H^2+124H

We move all terms to the left:

(H)-(-18H^2+124H)=0

We get rid of parentheses

18H^2-124H+H=0

We add all the numbers together, and all the variables

18H^2-123H=0

a = 18; b = -123; c = 0;
Δ = b2-4ac
Δ = -1232-4·18·0
Δ = 15129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{15129}=123$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-123)-123}{2*18}=\frac{0}{36}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-123)+123}{2*18}=\frac{246}{36}
=6+5/6
$