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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(x-2)

We get rid of parentheses

x-2

Back to the equation:

-(x-2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-3x-1=0

a = 1; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·1·(-1)
Δ = 13
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{13}}{2*1}=\frac{3-\sqrt{13}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{13}}{2*1}=\frac{3+\sqrt{13}}{2}
$