(3x-4)(4x-3)=(6x-4)

Solution for (3x-4)(4x-3)=(6x-4) equation:

(3x-4)(4x-3)=(6x-4)

We move all terms to the left:

(3x-4)(4x-3)-((6x-4))=0

We multiply parentheses ..

(+12x^2-9x-16x+12)-((6x-4))=0


We calculate terms in parentheses: -((6x-4)), so:

(6x-4)

We get rid of parentheses

6x-4

Back to the equation:

-(6x-4)


We get rid of parentheses

12x^2-9x-16x-6x+12+4=0

We add all the numbers together, and all the variables

12x^2-31x+16=0

a = 12; b = -31; c = +16;
Δ = b2-4ac
Δ = -312-4·12·16
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{193}}{2*12}=\frac{31-\sqrt{193}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{193}}{2*12}=\frac{31+\sqrt{193}}{24}
$