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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

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


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

(+x^2+3x+x+3)

We get rid of parentheses

x^2+3x+x+3

We add all the numbers together, and all the variables

x^2+4x+3

Back to the equation:

-(x^2+4x+3)


We add all the numbers together, and all the variables

-3x-(x^2+4x+3)-9=0

We get rid of parentheses

-x^2-3x-4x-3-9=0

We add all the numbers together, and all the variables

-1x^2-7x-12=0

a = -1; b = -7; c = -12;
Δ = b2-4ac
Δ = -72-4·(-1)·(-12)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*-1}=\frac{6}{-2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*-1}=\frac{8}{-2}
=-4
$