H(t)=16t^2+42t+5

Solution for H(t)=16t^2+42t+5 equation:

(H)=16H^2+42H+5

We move all terms to the left:

(H)-(16H^2+42H+5)=0

We get rid of parentheses

-16H^2+H-42H-5=0

We add all the numbers together, and all the variables

-16H^2-41H-5=0

a = -16; b = -41; c = -5;
Δ = b2-4ac
Δ = -412-4·(-16)·(-5)
Δ = 1361
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{-(-41)-\sqrt{1361}}{2*-16}=\frac{41-\sqrt{1361}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-41)+\sqrt{1361}}{2*-16}=\frac{41+\sqrt{1361}}{-32}
$