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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6n^2+n-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:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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