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

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

Simplifying
(2y2 + 6)(y2 + -1) = 0

Reorder the terms:
(6 + 2y2)(y2 + -1) = 0

Reorder the terms:
(6 + 2y2)(-1 + y2) = 0

Multiply (6 + 2y2) * (-1 + y2)
(6(-1 + y2) + 2y2 * (-1 + y2)) = 0
((-1 * 6 + y2 * 6) + 2y2 * (-1 + y2)) = 0
((-6 + 6y2) + 2y2 * (-1 + y2)) = 0
(-6 + 6y2 + (-1 * 2y2 + y2 * 2y2)) = 0
(-6 + 6y2 + (-2y2 + 2y4)) = 0

Combine like terms: 6y2 + -2y2 = 4y2
(-6 + 4y2 + 2y4) = 0

Solving
-6 + 4y2 + 2y4 = 0

Solving for variable 'y'.

Factor out the Greatest Common Factor (GCF), '2'.
2(-3 + 2y2 + y4) = 0

Factor a trinomial.
2((-3 + -1y2)(1 + -1y2)) = 0

Factor a difference between two squares.
2((-3 + -1y2)((1 + y)(1 + -1y))) = 0

Ignore the factor 2.
Subproblem 1
Set the factor '(-3 + -1y2)' equal to zero and attempt to solve:

Simplifying
-3 + -1y2 = 0

Solving
-3 + -1y2 = 0

Move all terms containing y to the left, all other terms to the right.

Add '3' to each side of the equation.
-3 + 3 + -1y2 = 0 + 3

Combine like terms: -3 + 3 = 0
0 + -1y2 = 0 + 3
-1y2 = 0 + 3

Combine like terms: 0 + 3 = 3
-1y2 = 3

Divide each side by '-1'.
y2 = -3

Simplifying
y2 = -3

The solution to this equation could not be determined.

This subproblem is being ignored because a solution could not be determined.
Subproblem 2
Set the factor '(1 + y)' equal to zero and attempt to solve:

Simplifying
1 + y = 0

Solving
1 + y = 0

Move all terms containing y to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + y = 0 + -1

Combine like terms: 1 + -1 = 0
0 + y = 0 + -1
y = 0 + -1

Combine like terms: 0 + -1 = -1
y = -1

Simplifying
y = -1
Subproblem 3
Set the factor '(1 + -1y)' equal to zero and attempt to solve:

Simplifying
1 + -1y = 0

Solving
1 + -1y = 0

Move all terms containing y to the left, all other terms to the right.

Add '-1' to each side of the equation.
1 + -1 + -1y = 0 + -1

Combine like terms: 1 + -1 = 0
0 + -1y = 0 + -1
-1y = 0 + -1

Combine like terms: 0 + -1 = -1
-1y = -1

Divide each side by '-1'.
y = 1

Simplifying
y = 1
Solution
y = {-1, 1}