0=3t^2-18t+24

Solution for 0=3t^2-18t+24 equation:

0=3t^2-18t+24

We move all terms to the left:

0-(3t^2-18t+24)=0

We add all the numbers together, and all the variables

-(3t^2-18t+24)=0

We get rid of parentheses

-3t^2+18t-24=0

a = -3; b = 18; c = -24;
Δ = b2-4ac
Δ = 182-4·(-3)·(-24)
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*-3}=\frac{-24}{-6}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*-3}=\frac{-12}{-6}
=+2
$