H=-16t^2+12t+42

Solution for H=-16t^2+12t+42 equation:

=-16H^2+12H+42

We move all terms to the left:

-(-16H^2+12H+42)=0

We get rid of parentheses

16H^2-12H-42=0

a = 16; b = -12; c = -42;
Δ = b2-4ac
Δ = -122-4·16·(-42)
Δ = 2832
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2832}=\sqrt{16*177}=\sqrt{16}*\sqrt{177}=4\sqrt{177}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{177}}{2*16}=\frac{12-4\sqrt{177}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{177}}{2*16}=\frac{12+4\sqrt{177}}{32}
$