(3y2-2)2=2(y2-2)+3y-8

Solution for (3y2-2)2=2(y2-2)+3y-8 equation:

(3y^2-2)2=2(y2-2)+3y-8

We move all terms to the left:

(3y^2-2)2-(2(y2-2)+3y-8)=0

We add all the numbers together, and all the variables

-(2(+y^2-2)+3y-8)+(3y^2-2)2=0

We multiply parentheses

-(2(+y^2-2)+3y-8)+6y^2-4=0


We calculate terms in parentheses: -(2(+y^2-2)+3y-8), so:

2(+y^2-2)+3y-8

We multiply parentheses

2y^2+3y-4-8

We add all the numbers together, and all the variables

2y^2+3y-12

Back to the equation:

-(2y^2+3y-12)


We get rid of parentheses

6y^2-2y^2-3y+12-4=0

We add all the numbers together, and all the variables

4y^2-3y+8=0

a = 4; b = -3; c = +8;
Δ = b2-4ac
Δ = -32-4·4·8
Δ = -119
Delta is less than zero, so there is no solution for the equation