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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-6x+2=0

a = 2; b = -6; c = +2;
Δ = b2-4ac
Δ = -62-4·2·2
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{5}}{2*2}=\frac{6-2\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{5}}{2*2}=\frac{6+2\sqrt{5}}{4}
$