3(y-4)+4(y+8)=10-5y(y-3)

Solution for 3(y-4)+4(y+8)=10-5y(y-3) equation:

3(y-4)+4(y+8)=10-5y(y-3)

We move all terms to the left:

3(y-4)+4(y+8)-(10-5y(y-3))=0

We multiply parentheses

3y+4y-(10-5y(y-3))-12+32=0


We calculate terms in parentheses: -(10-5y(y-3)), so:

10-5y(y-3)

determiningTheFunctionDomain
-5y(y-3)+10

We multiply parentheses

-5y^2+15y+10

Back to the equation:

-(-5y^2+15y+10)


We add all the numbers together, and all the variables

-(-5y^2+15y+10)+7y+20=0

We get rid of parentheses

5y^2-15y+7y-10+20=0

We add all the numbers together, and all the variables

5y^2-8y+10=0

a = 5; b = -8; c = +10;
Δ = b2-4ac
Δ = -82-4·5·10
Δ = -136
Delta is less than zero, so there is no solution for the equation