-3t^2+24t+1=0

Solution for -3t^2+24t+1=0 equation:

-3t^2+24t+1=0

a = -3; b = 24; c = +1;
Δ = b2-4ac
Δ = 242-4·(-3)·1
Δ = 588
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:

$\sqrt{\Delta}=\sqrt{588}=\sqrt{196*3}=\sqrt{196}*\sqrt{3}=14\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-14\sqrt{3}}{2*-3}=\frac{-24-14\sqrt{3}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+14\sqrt{3}}{2*-3}=\frac{-24+14\sqrt{3}}{-6}
$