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

Solution for 1(y-2)(y+2)=0 equation:

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

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

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

Multiply (-2 + y) * (2 + y)
1(-2(2 + y) + y(2 + y)) = 0
1((2 * -2 + y * -2) + y(2 + y)) = 0
1((-4 + -2y) + y(2 + y)) = 0
1(-4 + -2y + (2 * y + y * y)) = 0
1(-4 + -2y + (2y + y2)) = 0

Combine like terms: -2y + 2y = 0
1(-4 + 0 + y2) = 0
1(-4 + y2) = 0
(-4 * 1 + y2 * 1) = 0
(-4 + 1y2) = 0

Solving
-4 + 1y2 = 0

Solving for variable 'y'.

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

Add '4' to each side of the equation.
-4 + 4 + 1y2 = 0 + 4

Combine like terms: -4 + 4 = 0
0 + 1y2 = 0 + 4
1y2 = 0 + 4

Combine like terms: 0 + 4 = 4
1y2 = 4

Divide each side by '1'.
y2 = 4

Simplifying
y2 = 4

Take the square root of each side:
y = {-2, 2}