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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

x^2+x-3x-3

We add all the numbers together, and all the variables

x^2-2x-3

Back to the equation:

-(x^2-2x-3)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = 3; c = +3;
Δ = b2-4ac
Δ = 32-4·(-1)·3
Δ = 21
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{-(3)-\sqrt{21}}{2*-1}=\frac{-3-\sqrt{21}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{21}}{2*-1}=\frac{-3+\sqrt{21}}{-2}
$