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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

2x-((-3x+1)(x+1))-16=0

We multiply parentheses ..

-((-3x^2-3x+x+1))+2x-16=0


We calculate terms in parentheses: -((-3x^2-3x+x+1)), so:

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

We get rid of parentheses

-3x^2-3x+x+1

We add all the numbers together, and all the variables

-3x^2-2x+1

Back to the equation:

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


We get rid of parentheses

3x^2+2x+2x-1-16=0

We add all the numbers together, and all the variables

3x^2+4x-17=0

a = 3; b = 4; c = -17;
Δ = b2-4ac
Δ = 42-4·3·(-17)
Δ = 220
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{55}}{2*3}=\frac{-4-2\sqrt{55}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{55}}{2*3}=\frac{-4+2\sqrt{55}}{6}
$