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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+6x^2-10x-3x+5)2)+6=0


We calculate terms in parentheses: -((+6x^2-10x-3x+5)2), so:

(+6x^2-10x-3x+5)2

We multiply parentheses

12x^2-20x-6x+10

We add all the numbers together, and all the variables

12x^2-26x+10

Back to the equation:

-(12x^2-26x+10)


We get rid of parentheses

-12x^2+26x-10+6=0

We add all the numbers together, and all the variables

-12x^2+26x-4=0

a = -12; b = 26; c = -4;
Δ = b2-4ac
Δ = 262-4·(-12)·(-4)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-22}{2*-12}=\frac{-48}{-24}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+22}{2*-12}=\frac{-4}{-24}
=1/6
$