2r(3r-2)=0

Solution for 2r(3r-2)=0 equation:

2r(3r-2)=0

We multiply parentheses

6r^2-4r=0

a = 6; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·6·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*6}=\frac{0}{12}
=0
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*6}=\frac{8}{12}
=2/3
$