H=-3t^2+12t+42

Solution for H=-3t^2+12t+42 equation:

=-3H^2+12H+42

We move all terms to the left:

-(-3H^2+12H+42)=0

We get rid of parentheses

3H^2-12H-42=0

a = 3; b = -12; c = -42;
Δ = b2-4ac
Δ = -122-4·3·(-42)
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-18\sqrt{2}}{2*3}=\frac{12-18\sqrt{2}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+18\sqrt{2}}{2*3}=\frac{12+18\sqrt{2}}{6}
$