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

Solution for H(t)=-16t^2+42+0 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

16H^2+H-42=0

a = 16; b = 1; c = -42;
Δ = b2-4ac
Δ = 12-4·16·(-42)
Δ = 2689
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{-(1)-\sqrt{2689}}{2*16}=\frac{-1-\sqrt{2689}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{2689}}{2*16}=\frac{-1+\sqrt{2689}}{32}
$