0=-3t^2+22t-24

Solution for 0=-3t^2+22t-24 equation:

0=-3t^2+22t-24

We move all terms to the left:

0-(-3t^2+22t-24)=0

We add all the numbers together, and all the variables

-(-3t^2+22t-24)=0

We get rid of parentheses

3t^2-22t+24=0

a = 3; b = -22; c = +24;
Δ = b2-4ac
Δ = -222-4·3·24
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-14}{2*3}=\frac{8}{6}
=1+1/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+14}{2*3}=\frac{36}{6}
=6
$