0=-3t^2+24t-15

Solution for 0=-3t^2+24t-15 equation:

0=-3t^2+24t-15

We move all terms to the left:

0-(-3t^2+24t-15)=0

We add all the numbers together, and all the variables

-(-3t^2+24t-15)=0

We get rid of parentheses

3t^2-24t+15=0

a = 3; b = -24; c = +15;
Δ = b2-4ac
Δ = -242-4·3·15
Δ = 396
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6\sqrt{11}}{2*3}=\frac{24-6\sqrt{11}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6\sqrt{11}}{2*3}=\frac{24+6\sqrt{11}}{6}
$