2y(y-2)=(y+1)(y-6)-2

Solution for 2y(y-2)=(y+1)(y-6)-2 equation:

2y(y-2)=(y+1)(y-6)-2

We move all terms to the left:

2y(y-2)-((y+1)(y-6)-2)=0

We multiply parentheses

2y^2-4y-((y+1)(y-6)-2)=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+y^2-6y+y-6)-2), so:

(+y^2-6y+y-6)-2

We get rid of parentheses

y^2-6y+y-6-2

We add all the numbers together, and all the variables

y^2-5y-8

Back to the equation:

-(y^2-5y-8)


We add all the numbers together, and all the variables

2y^2-4y-(y^2-5y-8)=0

We get rid of parentheses

2y^2-y^2-4y+5y+8=0

We add all the numbers together, and all the variables

y^2+y+8=0

a = 1; b = 1; c = +8;
Δ = b2-4ac
Δ = 12-4·1·8
Δ = -31
Delta is less than zero, so there is no solution for the equation