3t^2-24t+8=0

Solution for 3t^2-24t+8=0 equation:

3t^2-24t+8=0

a = 3; b = -24; c = +8;
Δ = b2-4ac
Δ = -242-4·3·8
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{30}}{2*3}=\frac{24-4\sqrt{30}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{30}}{2*3}=\frac{24+4\sqrt{30}}{6}
$