H(t)=3t^2+12t-2

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

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

We move all terms to the left:

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

We get rid of parentheses

-3H^2+H-12H+2=0

We add all the numbers together, and all the variables

-3H^2-11H+2=0

a = -3; b = -11; c = +2;
Δ = b2-4ac
Δ = -112-4·(-3)·2
Δ = 145
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{-(-11)-\sqrt{145}}{2*-3}=\frac{11-\sqrt{145}}{-6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{145}}{2*-3}=\frac{11+\sqrt{145}}{-6}
$