40=(3x-5)(2x-1)(2)

Solution for 40=(3x-5)(2x-1)(2) equation:

40=(3x-5)(2x-1)(2)

We move all terms to the left:

40-((3x-5)(2x-1)(2))=0

We multiply parentheses ..

-((+6x^2-3x-10x+5)2)+40=0


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

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

We multiply parentheses

12x^2-6x-20x+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+40=0

We add all the numbers together, and all the variables

-12x^2+26x+30=0

a = -12; b = 26; c = +30;
Δ = b2-4ac
Δ = 262-4·(-12)·30
Δ = 2116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2116}=46$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-46}{2*-12}=\frac{-72}{-24}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+46}{2*-12}=\frac{20}{-24}
=-5/6
$