(3x-1)(4x-3)=(8x-7)(3x-1)

Solution for (3x-1)(4x-3)=(8x-7)(3x-1) equation:

(3x-1)(4x-3)=(8x-7)(3x-1)

We move all terms to the left:

(3x-1)(4x-3)-((8x-7)(3x-1))=0

We multiply parentheses ..

(+12x^2-9x-4x+3)-((8x-7)(3x-1))=0


We calculate terms in parentheses: -((8x-7)(3x-1)), so:

(8x-7)(3x-1)

We multiply parentheses ..

(+24x^2-8x-21x+7)

We get rid of parentheses

24x^2-8x-21x+7

We add all the numbers together, and all the variables

24x^2-29x+7

Back to the equation:

-(24x^2-29x+7)


We get rid of parentheses

12x^2-24x^2-9x-4x+29x+3-7=0

We add all the numbers together, and all the variables

-12x^2+16x-4=0

a = -12; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·(-12)·(-4)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*-12}=\frac{-24}{-24}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*-12}=\frac{-8}{-24}
=1/3
$