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

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

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

a = -3; b = 6; c = +2;
Δ = b2-4ac
Δ = 62-4·(-3)·2
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{15}}{2*-3}=\frac{-6-2\sqrt{15}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{15}}{2*-3}=\frac{-6+2\sqrt{15}}{-6}
$