y-6=1/2(2y-20)(3y-4)

Solution for y-6=1/2(2y-20)(3y-4) equation:

y-6=1/2(2y-20)(3y-4)

We move all terms to the left:

y-6-(1/2(2y-20)(3y-4))=0


Domain of the equation: 2(2y-20)(3y-4))!=0

y∈R


We multiply parentheses ..

-(1/2(+6y^2-8y-60y+80))+y-6=0

We multiply all the terms by the denominator


-(1+y*2(+6y^2-8y-60y+80))-6*2(+6y^2-8y-60y+80))=0


We calculate terms in parentheses: -(1+y*2(+6y^2-8y-60y+80)), so:

1+y*2(+6y^2-8y-60y+80)

determiningTheFunctionDomain
y*2(+6y^2-8y-60y+80)+1

Wy multiply elements

2y^2(++1

We use the square of the difference formula

2y^2(+1

Back to the equation:

-(2y^2(+1)


We add all the numbers together, and all the variables

-(2y^21-6*2(+6y^2-8y-60y+80))=0


We calculate terms in parentheses: -(2y^21-6*2(+6y^2-8y-60y+80)), so:

2y^21-6*2(+6y^2-8y-60y+80)

We do not support eypression: y^21