3t^2-30t+24=0

Solution for 3t^2-30t+24=0 equation:

3t^2-30t+24=0

a = 3; b = -30; c = +24;
Δ = b2-4ac
Δ = -302-4·3·24
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{17}}{2*3}=\frac{30-6\sqrt{17}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{17}}{2*3}=\frac{30+6\sqrt{17}}{6}
$