1.5t^2-24t+24=0

Solution for 1.5t^2-24t+24=0 equation:

1.5t^2-24t+24=0

a = 1.5; b = -24; c = +24;
Δ = b2-4ac
Δ = -242-4·1.5·24
Δ = 432
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:

$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12\sqrt{3}}{2*1.5}=\frac{24-12\sqrt{3}}{3}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12\sqrt{3}}{2*1.5}=\frac{24+12\sqrt{3}}{3}
$