H(t)=16t^2+40t+3

Solution for H(t)=16t^2+40t+3 equation:

(H)=16H^2+40H+3

We move all terms to the left:

(H)-(16H^2+40H+3)=0

We get rid of parentheses

-16H^2+H-40H-3=0

We add all the numbers together, and all the variables

-16H^2-39H-3=0

a = -16; b = -39; c = -3;
Δ = b2-4ac
Δ = -392-4·(-16)·(-3)
Δ = 1329
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{-(-39)-\sqrt{1329}}{2*-16}=\frac{39-\sqrt{1329}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1329}}{2*-16}=\frac{39+\sqrt{1329}}{-32}
$