0=12t^2-18t+6

Solution for 0=12t^2-18t+6 equation:

0=12t^2-18t+6

We move all terms to the left:

0-(12t^2-18t+6)=0

We add all the numbers together, and all the variables

-(12t^2-18t+6)=0

We get rid of parentheses

-12t^2+18t-6=0

a = -12; b = 18; c = -6;
Δ = b2-4ac
Δ = 182-4·(-12)·(-6)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*-12}=\frac{-24}{-24}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*-12}=\frac{-12}{-24}
=1/2
$