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

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



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

a = -3; b = -3; c = +2;
Δ = b2-4ac
Δ = -32-4·(-3)·2
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{33}}{2*-3}=\frac{3-\sqrt{33}}{-6}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*-3}=\frac{3+\sqrt{33}}{-6}
$

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