H(t)=16t^2+144t+5

Solution for H(t)=16t^2+144t+5 equation:

(H)=16H^2+144H+5

We move all terms to the left:

(H)-(16H^2+144H+5)=0

We get rid of parentheses

-16H^2+H-144H-5=0

We add all the numbers together, and all the variables

-16H^2-143H-5=0

a = -16; b = -143; c = -5;
Δ = b2-4ac
Δ = -1432-4·(-16)·(-5)
Δ = 20129
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{-(-143)-\sqrt{20129}}{2*-16}=\frac{143-\sqrt{20129}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-143)+\sqrt{20129}}{2*-16}=\frac{143+\sqrt{20129}}{-32}
$