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

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

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

We move all terms to the left:

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


Domain of the equation: (3x-5)(2x-1))!=0

x∈R


We multiply parentheses ..

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

We multiply all the terms by the denominator


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


We calculate terms in parentheses: -(40+2*(+6x^2-3x-10x+5)), so:

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

determiningTheFunctionDomain
2*(+6x^2-3x-10x+5)+40

We multiply parentheses

12x^2-6x-20x+10+40

We add all the numbers together, and all the variables

12x^2-26x+50

Back to the equation:

-(12x^2-26x+50)


We get rid of parentheses

-12x^2+26x-50=0

a = -12; b = 26; c = -50;
Δ = b2-4ac
Δ = 262-4·(-12)·(-50)
Δ = -1724
Delta is less than zero, so there is no solution for the equation