H(t)=-16t^2+32t+21

Solution for H(t)=-16t^2+32t+21 equation:

(H)=-16H^2+32H+21

We move all terms to the left:

(H)-(-16H^2+32H+21)=0

We get rid of parentheses

16H^2-32H+H-21=0

We add all the numbers together, and all the variables

16H^2-31H-21=0

a = 16; b = -31; c = -21;
Δ = b2-4ac
Δ = -312-4·16·(-21)
Δ = 2305
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{-(-31)-\sqrt{2305}}{2*16}=\frac{31-\sqrt{2305}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{2305}}{2*16}=\frac{31+\sqrt{2305}}{32}
$