6(2x-1)-2x(3x-4)=5(3-4x)

Solution for 6(2x-1)-2x(3x-4)=5(3-4x) equation:

6(2x-1)-2x(3x-4)=5(3-4x)

We move all terms to the left:

6(2x-1)-2x(3x-4)-(5(3-4x))=0

We add all the numbers together, and all the variables

6(2x-1)-2x(3x-4)-(5(-4x+3))=0

We multiply parentheses

-6x^2+12x+8x-(5(-4x+3))-6=0


We calculate terms in parentheses: -(5(-4x+3)), so:

5(-4x+3)

We multiply parentheses

-20x+15

Back to the equation:

-(-20x+15)


We add all the numbers together, and all the variables

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

We get rid of parentheses

-6x^2+20x+20x-15-6=0

We add all the numbers together, and all the variables

-6x^2+40x-21=0

a = -6; b = 40; c = -21;
Δ = b2-4ac
Δ = 402-4·(-6)·(-21)
Δ = 1096
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:

$\sqrt{\Delta}=\sqrt{1096}=\sqrt{4*274}=\sqrt{4}*\sqrt{274}=2\sqrt{274}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{274}}{2*-6}=\frac{-40-2\sqrt{274}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{274}}{2*-6}=\frac{-40+2\sqrt{274}}{-12}
$