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

Solution for 0=(3x+1)(2x-6) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+6x^2-18x+2x-6))=0


We calculate terms in parentheses: -((+6x^2-18x+2x-6)), so:

(+6x^2-18x+2x-6)

We get rid of parentheses

6x^2-18x+2x-6

We add all the numbers together, and all the variables

6x^2-16x-6

Back to the equation:

-(6x^2-16x-6)


We get rid of parentheses

-6x^2+16x+6=0

a = -6; b = 16; c = +6;
Δ = b2-4ac
Δ = 162-4·(-6)·6
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-20}{2*-6}=\frac{-36}{-12}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+20}{2*-6}=\frac{4}{-12}
=-1/3
$