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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We use the square of the difference formula

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-3x+1=0

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