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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+4x^2-10x-2x+5)-((-2x+1)(x+1))=0


We calculate terms in parentheses: -((-2x+1)(x+1)), so:

(-2x+1)(x+1)

We multiply parentheses ..

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

We get rid of parentheses

-2x^2-2x+x+1

We add all the numbers together, and all the variables

-2x^2-1x+1

Back to the equation:

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


We get rid of parentheses

4x^2+2x^2-10x-2x+1x+5-1=0

We add all the numbers together, and all the variables

6x^2-11x+4=0

a = 6; b = -11; c = +4;
Δ = b2-4ac
Δ = -112-4·6·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*6}=\frac{6}{12}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*6}=\frac{16}{12}
=1+1/3
$