H=-16t^2+4t+42

Solution for H=-16t^2+4t+42 equation:

=-16H^2+4H+42

We move all terms to the left:

-(-16H^2+4H+42)=0

We get rid of parentheses

16H^2-4H-42=0

a = 16; b = -4; c = -42;
Δ = b2-4ac
Δ = -42-4·16·(-42)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-52}{2*16}=\frac{-48}{32}
=-1+1/2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+52}{2*16}=\frac{56}{32}
=1+3/4
$