H(t)=-16t^2+11t+33

Solution for H(t)=-16t^2+11t+33 equation:

(H)=-16H^2+11H+33

We move all terms to the left:

(H)-(-16H^2+11H+33)=0

We get rid of parentheses

16H^2-11H+H-33=0

We add all the numbers together, and all the variables

16H^2-10H-33=0

a = 16; b = -10; c = -33;
Δ = b2-4ac
Δ = -102-4·16·(-33)
Δ = 2212
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:

$\sqrt{\Delta}=\sqrt{2212}=\sqrt{4*553}=\sqrt{4}*\sqrt{553}=2\sqrt{553}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{553}}{2*16}=\frac{10-2\sqrt{553}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{553}}{2*16}=\frac{10+2\sqrt{553}}{32}
$