0=9t^2-18t^2+6t

Solution for 0=9t^2-18t^2+6t equation:

0=9t^2-18t^2+6t

We move all terms to the left:

0-(9t^2-18t^2+6t)=0

We add all the numbers together, and all the variables

-(9t^2-18t^2+6t)=0

We get rid of parentheses

-9t^2+18t^2-6t=0

We add all the numbers together, and all the variables

9t^2-6t=0

a = 9; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·9·0
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{-(-6)-6}{2*9}=\frac{0}{18}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*9}=\frac{12}{18}
=2/3
$