68=((3x-2)(x-6))

Solution for 68=((3x-2)(x-6)) equation:

68=((3x-2)(x-6))

We move all terms to the left:

68-(((3x-2)(x-6)))=0

We multiply parentheses ..

-(((+3x^2-18x-2x+12)))+68=0


We calculate terms in parentheses: -(((+3x^2-18x-2x+12))), so:

((+3x^2-18x-2x+12))


We calculate terms in parentheses: +((+3x^2-18x-2x+12)), so:

(+3x^2-18x-2x+12)

We get rid of parentheses

3x^2-18x-2x+12

We add all the numbers together, and all the variables

3x^2-20x+12

Back to the equation:

+(3x^2-20x+12)


We get rid of parentheses

3x^2-20x+12

Back to the equation:

-(3x^2-20x+12)


We get rid of parentheses

-3x^2+20x-12+68=0

We add all the numbers together, and all the variables

-3x^2+20x+56=0

a = -3; b = 20; c = +56;
Δ = b2-4ac
Δ = 202-4·(-3)·56
Δ = 1072
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:

$\sqrt{\Delta}=\sqrt{1072}=\sqrt{16*67}=\sqrt{16}*\sqrt{67}=4\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{67}}{2*-3}=\frac{-20-4\sqrt{67}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{67}}{2*-3}=\frac{-20+4\sqrt{67}}{-6}
$