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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

3(x+2)

We multiply parentheses

3x+6

Back to the equation:

-(3x+6)


We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-2x-8=0

a = 6; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·6·(-8)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*6}=\frac{-12}{12}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*6}=\frac{16}{12}
=1+1/3
$