6x-12=3(x-3)(x-4)

Solution for 6x-12=3(x-3)(x-4) equation:

6x-12=3(x-3)(x-4)

We move all terms to the left:

6x-12-(3(x-3)(x-4))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -(3(+x^2-4x-3x+12)), so:

3(+x^2-4x-3x+12)

We multiply parentheses

3x^2-12x-9x+36

We add all the numbers together, and all the variables

3x^2-21x+36

Back to the equation:

-(3x^2-21x+36)


We add all the numbers together, and all the variables

6x-(3x^2-21x+36)-12=0

We get rid of parentheses

-3x^2+6x+21x-36-12=0

We add all the numbers together, and all the variables

-3x^2+27x-48=0

a = -3; b = 27; c = -48;
Δ = b2-4ac
Δ = 272-4·(-3)·(-48)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3\sqrt{17}}{2*-3}=\frac{-27-3\sqrt{17}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3\sqrt{17}}{2*-3}=\frac{-27+3\sqrt{17}}{-6}
$