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

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

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

a = -3; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-3)·3
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{10}}{2*-3}=\frac{-2-2\sqrt{10}}{-6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{10}}{2*-3}=\frac{-2+2\sqrt{10}}{-6}
$