3/4(20y-8)+5=1/2(20y+8)

Solution for 3/4(20y-8)+5=1/2(20y+8) equation:

3/4(20y-8)+5=1/2(20y+8)

We move all terms to the left:

3/4(20y-8)+5-(1/2(20y+8))=0


Domain of the equation: 4(20y-8)!=0

y∈R



Domain of the equation: 2(20y+8))!=0

y∈R


We calculate fractions


(6y2/(4(20y-8)*2(20y+8)))+(-4y2/(4(20y-8)*2(20y+8)))+5=0


We calculate terms in parentheses: +(6y2/(4(20y-8)*2(20y+8))), so:

6y2/(4(20y-8)*2(20y+8))

We multiply all the terms by the denominator


6y2

We add all the numbers together, and all the variables

6y^2

Back to the equation:

+(6y^2)



We calculate terms in parentheses: +(-4y2/(4(20y-8)*2(20y+8))), so:

-4y2/(4(20y-8)*2(20y+8))

We multiply all the terms by the denominator


-4y2

We add all the numbers together, and all the variables

-4y^2

Back to the equation:

+(-4y^2)


We add all the numbers together, and all the variables

6y^2+(-4y^2)+5=0

We get rid of parentheses

6y^2-4y^2+5=0

We add all the numbers together, and all the variables

2y^2+5=0

a = 2; b = 0; c = +5;
Δ = b2-4ac
Δ = 02-4·2·5
Δ = -40
Delta is less than zero, so there is no solution for the equation