H=-3t2+12t+1

Solution for H=-3t2+12t+1 equation:

=-3H^2+12H+1

We move all terms to the left:

-(-3H^2+12H+1)=0

We get rid of parentheses

3H^2-12H-1=0

a = 3; b = -12; c = -1;
Δ = b2-4ac
Δ = -122-4·3·(-1)
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{39}}{2*3}=\frac{12-2\sqrt{39}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{39}}{2*3}=\frac{12+2\sqrt{39}}{6}
$