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

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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 $

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