H(t)=9t^2-24t+16

Solution for H(t)=9t^2-24t+16 equation:

(H)=9H^2-24H+16

We move all terms to the left:

(H)-(9H^2-24H+16)=0

We get rid of parentheses

-9H^2+H+24H-16=0

We add all the numbers together, and all the variables

-9H^2+25H-16=0

a = -9; b = 25; c = -16;
Δ = b2-4ac
Δ = 252-4·(-9)·(-16)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-7}{2*-9}=\frac{-32}{-18}
=1+7/9
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+7}{2*-9}=\frac{-18}{-18}
=1
$