8y-4+3(y+7)=6y-y(y-3)

Solution for 8y-4+3(y+7)=6y-y(y-3) equation:

8y-4+3(y+7)=6y-y(y-3)

We move all terms to the left:

8y-4+3(y+7)-(6y-y(y-3))=0

We multiply parentheses

8y+3y-(6y-y(y-3))+21-4=0


We calculate terms in parentheses: -(6y-y(y-3)), so:

6y-y(y-3)

We multiply parentheses

-y^2+6y+3y

We add all the numbers together, and all the variables

-1y^2+9y

Back to the equation:

-(-1y^2+9y)


We add all the numbers together, and all the variables

-(-1y^2+9y)+11y+17=0

We get rid of parentheses

1y^2-9y+11y+17=0

We add all the numbers together, and all the variables

y^2+2y+17=0

a = 1; b = 2; c = +17;
Δ = b2-4ac
Δ = 22-4·1·17
Δ = -64
Delta is less than zero, so there is no solution for the equation