(4x-3)(3x-4)=(2-5)(6x-4)

Solution for (4x-3)(3x-4)=(2-5)(6x-4) equation:

(4x-3)(3x-4)=(2-5)(6x-4)

We move all terms to the left:

(4x-3)(3x-4)-((2-5)(6x-4))=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+12x^2-16x-9x+12)-((-3)(6x-4))=0


We calculate terms in parentheses: -((-3)(6x-4)), so:

(-3)(6x-4)

We multiply parentheses ..

(-18x+12)

We get rid of parentheses

-18x+12

Back to the equation:

-(-18x+12)


We get rid of parentheses

12x^2-16x-9x+18x+12-12=0

We add all the numbers together, and all the variables

12x^2-7x=0

a = 12; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·12·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*12}=\frac{0}{24}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*12}=\frac{14}{24}
=7/12
$