0=3t^2-18t+15

Solution for 0=3t^2-18t+15 equation:

0=3t^2-18t+15

We move all terms to the left:

0-(3t^2-18t+15)=0

We add all the numbers together, and all the variables

-(3t^2-18t+15)=0

We get rid of parentheses

-3t^2+18t-15=0

a = -3; b = 18; c = -15;
Δ = b2-4ac
Δ = 182-4·(-3)·(-15)
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{-(18)-12}{2*-3}=\frac{-30}{-6}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12}{2*-3}=\frac{-6}{-6}
=1
$