0=18t^2+9t

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

0=18t^2+9t

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-18t^2-9t=0

a = -18; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·(-18)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*-18}=\frac{0}{-36}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*-18}=\frac{18}{-36}
=-1/2
$