3x(7x-4)=(3x-4)(5x-2)

Solution for 3x(7x-4)=(3x-4)(5x-2) equation:

3x(7x-4)=(3x-4)(5x-2)

We move all terms to the left:

3x(7x-4)-((3x-4)(5x-2))=0

We multiply parentheses

21x^2-12x-((3x-4)(5x-2))=0

We multiply parentheses ..

21x^2-((+15x^2-6x-20x+8))-12x=0


We calculate terms in parentheses: -((+15x^2-6x-20x+8)), so:

(+15x^2-6x-20x+8)

We get rid of parentheses

15x^2-6x-20x+8

We add all the numbers together, and all the variables

15x^2-26x+8

Back to the equation:

-(15x^2-26x+8)


We add all the numbers together, and all the variables

21x^2-12x-(15x^2-26x+8)=0

We get rid of parentheses

21x^2-15x^2-12x+26x-8=0

We add all the numbers together, and all the variables

6x^2+14x-8=0

a = 6; b = 14; c = -8;
Δ = b2-4ac
Δ = 142-4·6·(-8)
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{97}}{2*6}=\frac{-14-2\sqrt{97}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{97}}{2*6}=\frac{-14+2\sqrt{97}}{12}
$