0=3t^2+12t

Solution for 0=3t^2+12t equation:

0=3t^2+12t

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-3t^2-12t=0

a = -3; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·(-3)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*-3}=\frac{0}{-6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*-3}=\frac{24}{-6}
=-4
$