(x+1)(3x-1)=(2x-1)(4x-2)

Solution for (x+1)(3x-1)=(2x-1)(4x-2) equation:

(x+1)(3x-1)=(2x-1)(4x-2)

We move all terms to the left:

(x+1)(3x-1)-((2x-1)(4x-2))=0

We multiply parentheses ..

(+3x^2-1x+3x-1)-((2x-1)(4x-2))=0


We calculate terms in parentheses: -((2x-1)(4x-2)), so:

(2x-1)(4x-2)

We multiply parentheses ..

(+8x^2-4x-4x+2)

We get rid of parentheses

8x^2-4x-4x+2

We add all the numbers together, and all the variables

8x^2-8x+2

Back to the equation:

-(8x^2-8x+2)


We get rid of parentheses

3x^2-8x^2-1x+3x+8x-1-2=0

We add all the numbers together, and all the variables

-5x^2+10x-3=0

a = -5; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·(-5)·(-3)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{10}}{2*-5}=\frac{-10-2\sqrt{10}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{10}}{2*-5}=\frac{-10+2\sqrt{10}}{-10}
$