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

Solution for H(t)=3t^2-12t+0 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3H^2+13H=0

a = -3; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-3)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-3}=\frac{-26}{-6}
=4+1/3
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-3}=\frac{0}{-6}
=0
$