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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+6x^2-2x+3x-1))=0


We calculate terms in parentheses: -((+6x^2-2x+3x-1)), so:

(+6x^2-2x+3x-1)

We get rid of parentheses

6x^2-2x+3x-1

We add all the numbers together, and all the variables

6x^2+x-1

Back to the equation:

-(6x^2+x-1)


We get rid of parentheses

-6x^2-x+1=0

We add all the numbers together, and all the variables

-6x^2-1x+1=0

a = -6; b = -1; c = +1;
Δ = b2-4ac
Δ = -12-4·(-6)·1
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*-6}=\frac{-4}{-12}
=1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*-6}=\frac{6}{-12}
=-1/2
$