-3q^2-2q+1=0

Solution for -3q^2-2q+1=0 equation:

-3q^2-2q+1=0

a = -3; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-3)·1
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*-3}=\frac{-2}{-6}
=1/3
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*-3}=\frac{6}{-6}
=-1
$