(2x-4)(3x+1)=12

Solution for (2x-4)(3x+1)=12 equation:

(2x-4)(3x+1)=12

We move all terms to the left:

(2x-4)(3x+1)-(12)=0

We multiply parentheses ..

(+6x^2+2x-12x-4)-12=0

We get rid of parentheses

6x^2+2x-12x-4-12=0

We add all the numbers together, and all the variables

6x^2-10x-16=0

a = 6; b = -10; c = -16;
Δ = b2-4ac
Δ = -102-4·6·(-16)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*6}=\frac{-12}{12}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*6}=\frac{32}{12}
=2+2/3
$