H(t)=-16t^2+21t+5

Solution for H(t)=-16t^2+21t+5 equation:

(H)=-16H^2+21H+5

We move all terms to the left:

(H)-(-16H^2+21H+5)=0

We get rid of parentheses

16H^2-21H+H-5=0

We add all the numbers together, and all the variables

16H^2-20H-5=0

a = 16; b = -20; c = -5;
Δ = b2-4ac
Δ = -202-4·16·(-5)
Δ = 720
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:

$\sqrt{\Delta}=\sqrt{720}=\sqrt{144*5}=\sqrt{144}*\sqrt{5}=12\sqrt{5}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-12\sqrt{5}}{2*16}=\frac{20-12\sqrt{5}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+12\sqrt{5}}{2*16}=\frac{20+12\sqrt{5}}{32}
$