(3x+1)(2x-1)-2x=(2x-3)+6x+5

Solution for (3x+1)(2x-1)-2x=(2x-3)+6x+5 equation:

(3x+1)(2x-1)-2x=(2x-3)+6x+5

We move all terms to the left:

(3x+1)(2x-1)-2x-((2x-3)+6x+5)=0

We add all the numbers together, and all the variables

-2x+(3x+1)(2x-1)-((2x-3)+6x+5)=0

We multiply parentheses ..

(+6x^2-3x+2x-1)-2x-((2x-3)+6x+5)=0


We calculate terms in parentheses: -((2x-3)+6x+5), so:

(2x-3)+6x+5

We add all the numbers together, and all the variables

6x+(2x-3)+5

We get rid of parentheses

6x+2x-3+5

We add all the numbers together, and all the variables

8x+2

Back to the equation:

-(8x+2)


We get rid of parentheses

6x^2-3x+2x-2x-8x-1-2=0

We add all the numbers together, and all the variables

6x^2-11x-3=0

a = 6; b = -11; c = -3;
Δ = b2-4ac
Δ = -112-4·6·(-3)
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{193}}{2*6}=\frac{11-\sqrt{193}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{193}}{2*6}=\frac{11+\sqrt{193}}{12}
$