H(t)=6+16t-3t^2

Solution for H(t)=6+16t-3t^2 equation:

(H)=6+16H-3H^2

We move all terms to the left:

(H)-(6+16H-3H^2)=0

We get rid of parentheses

3H^2-16H+H-6=0

We add all the numbers together, and all the variables

3H^2-15H-6=0

a = 3; b = -15; c = -6;
Δ = b2-4ac
Δ = -152-4·3·(-6)
Δ = 297
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{33}}{2*3}=\frac{15-3\sqrt{33}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{33}}{2*3}=\frac{15+3\sqrt{33}}{6}
$