0=15t^2-18t

Solution for 0=15t^2-18t equation:

0=15t^2-18t

We move all terms to the left:

0-(15t^2-18t)=0

We add all the numbers together, and all the variables

-(15t^2-18t)=0

We get rid of parentheses

-15t^2+18t=0

a = -15; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·(-15)·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*-15}=\frac{-36}{-30}
=1+1/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*-15}=\frac{0}{-30}
=0
$