(3c-2)(2c+1)=0

Solution for (3c-2)(2c+1)=0 equation:

(3c-2)(2c+1)=0

We multiply parentheses ..

(+6c^2+3c-4c-2)=0

We get rid of parentheses

6c^2+3c-4c-2=0

We add all the numbers together, and all the variables

6c^2-1c-2=0

a = 6; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·6·(-2)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*6}=\frac{-6}{12}
=-1/2
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*6}=\frac{8}{12}
=2/3
$