-3r^2-2r+2=0

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

-3r^2-2r+2=0

a = -3; b = -2; c = +2;
Δ = b2-4ac
Δ = -22-4·(-3)·2
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-3}=\frac{2-2\sqrt{7}}{-6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-3}=\frac{2+2\sqrt{7}}{-6}
$