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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

4(3x-4)x

We multiply parentheses

12x^2-16x

Back to the equation:

-(12x^2-16x)


We get rid of parentheses

-12x^2+6x+16x-3=0

We add all the numbers together, and all the variables

-12x^2+22x-3=0

a = -12; b = 22; c = -3;
Δ = b2-4ac
Δ = 222-4·(-12)·(-3)
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{85}}{2*-12}=\frac{-22-2\sqrt{85}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{85}}{2*-12}=\frac{-22+2\sqrt{85}}{-24}
$