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

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

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

We add all the numbers together, and all the variables

3t^2-2t=0

a = 3; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·3·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*3}=\frac{0}{6}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*3}=\frac{4}{6}
=2/3
$