(2x-1)(x-3)=(x-5)(x-1)

Solution for (2x-1)(x-3)=(x-5)(x-1) equation:

(2x-1)(x-3)=(x-5)(x-1)

We move all terms to the left:

(2x-1)(x-3)-((x-5)(x-1))=0

We multiply parentheses ..

(+2x^2-6x-1x+3)-((x-5)(x-1))=0


We calculate terms in parentheses: -((x-5)(x-1)), so:

(x-5)(x-1)

We multiply parentheses ..

(+x^2-1x-5x+5)

We get rid of parentheses

x^2-1x-5x+5

We add all the numbers together, and all the variables

x^2-6x+5

Back to the equation:

-(x^2-6x+5)


We get rid of parentheses

2x^2-x^2-6x-1x+6x+3-5=0

We add all the numbers together, and all the variables

x^2-1x-2=0

a = 1; b = -1; c = -2;
Δ = b2-4ac
Δ = -12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-3}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+3}{2*1}=\frac{4}{2}
=2
$