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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16H^2-1H-33=0

a = 16; b = -1; c = -33;
Δ = b2-4ac
Δ = -12-4·16·(-33)
Δ = 2113
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{-(-1)-\sqrt{2113}}{2*16}=\frac{1-\sqrt{2113}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{2113}}{2*16}=\frac{1+\sqrt{2113}}{32}
$