-3t^2-12t+6=0

Solution for -3t^2-12t+6=0 equation:

-3t^2-12t+6=0

a = -3; b = -12; c = +6;
Δ = b2-4ac
Δ = -122-4·(-3)·6
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{6}}{2*-3}=\frac{12-6\sqrt{6}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{6}}{2*-3}=\frac{12+6\sqrt{6}}{-6}
$