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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+6x^2+9x-2x-3))=0


We calculate terms in parentheses: -((+6x^2+9x-2x-3)), so:

(+6x^2+9x-2x-3)

We get rid of parentheses

6x^2+9x-2x-3

We add all the numbers together, and all the variables

6x^2+7x-3

Back to the equation:

-(6x^2+7x-3)


We get rid of parentheses

-6x^2-7x+3=0

a = -6; b = -7; c = +3;
Δ = b2-4ac
Δ = -72-4·(-6)·3
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-11}{2*-6}=\frac{-4}{-12}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+11}{2*-6}=\frac{18}{-12}
=-1+1/2
$