H(t)=6+16t-2t^2

Solution for H(t)=6+16t-2t^2 equation:

(H)=6+16H-2H^2

We move all terms to the left:

(H)-(6+16H-2H^2)=0

We get rid of parentheses

2H^2-16H+H-6=0

We add all the numbers together, and all the variables

2H^2-15H-6=0

a = 2; b = -15; c = -6;
Δ = b2-4ac
Δ = -152-4·2·(-6)
Δ = 273
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{273}}{2*2}=\frac{15-\sqrt{273}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{273}}{2*2}=\frac{15+\sqrt{273}}{4}
$