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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2H^2-15H-16=0

a = 2; b = -15; c = -16;
Δ = b2-4ac
Δ = -152-4·2·(-16)
Δ = 353
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{-(-15)-\sqrt{353}}{2*2}=\frac{15-\sqrt{353}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{353}}{2*2}=\frac{15+\sqrt{353}}{4}
$