1/5(20x+20)-20=1/3(24x-12)

Solution for 1/5(20x+20)-20=1/3(24x-12) equation:

1/5(20x+20)-20=1/3(24x-12)

We move all terms to the left:

1/5(20x+20)-20-(1/3(24x-12))=0


Domain of the equation: 5(20x+20)!=0

x∈R



Domain of the equation: 3(24x-12))!=0

x∈R


We calculate fractions


(3x2/(5(20x+20)*3(24x-12)))+(-5x2/(5(20x+20)*3(24x-12)))-20=0


We calculate terms in parentheses: +(3x2/(5(20x+20)*3(24x-12))), so:

3x2/(5(20x+20)*3(24x-12))

We multiply all the terms by the denominator


3x2

We add all the numbers together, and all the variables

3x^2

Back to the equation:

+(3x^2)



We calculate terms in parentheses: +(-5x2/(5(20x+20)*3(24x-12))), so:

-5x2/(5(20x+20)*3(24x-12))

We multiply all the terms by the denominator


-5x2

We add all the numbers together, and all the variables

-5x^2

Back to the equation:

+(-5x^2)


We add all the numbers together, and all the variables

3x^2+(-5x^2)-20=0

We get rid of parentheses

3x^2-5x^2-20=0

We add all the numbers together, and all the variables

-2x^2-20=0

a = -2; b = 0; c = -20;
Δ = b2-4ac
Δ = 02-4·(-2)·(-20)
Δ = -160
Delta is less than zero, so there is no solution for the equation