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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((x-1)), so:

(x-1)

We get rid of parentheses

x-1

Back to the equation:

-(x-1)


We get rid of parentheses

-9x^2+3x-3x-6x-x+1+1+2=0

We add all the numbers together, and all the variables

-9x^2-7x+4=0

a = -9; b = -7; c = +4;
Δ = b2-4ac
Δ = -72-4·(-9)·4
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{-(-7)-\sqrt{193}}{2*-9}=\frac{7-\sqrt{193}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{193}}{2*-9}=\frac{7+\sqrt{193}}{-18}
$