(3x+1)(2x-3)-2(x+1)=7x+1

Solution for (3x+1)(2x-3)-2(x+1)=7x+1 equation:

(3x+1)(2x-3)-2(x+1)=7x+1

We move all terms to the left:

(3x+1)(2x-3)-2(x+1)-(7x+1)=0

We multiply parentheses

(3x+1)(2x-3)-2x-(7x+1)-2=0

We get rid of parentheses

(3x+1)(2x-3)-2x-7x-1-2=0

We multiply parentheses ..

(+6x^2-9x+2x-3)-2x-7x-1-2=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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