2y^2+128=40y^2

Solution for 2y^2+128=40y^2 equation:

Simplifying
2y2 + 128 = 40y2

Reorder the terms:
128 + 2y2 = 40y2

Solving
128 + 2y2 = 40y2

Solving for variable 'y'.

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

Add '-40y2' to each side of the equation.
128 + 2y2 + -40y2 = 40y2 + -40y2

Combine like terms: 2y2 + -40y2 = -38y2
128 + -38y2 = 40y2 + -40y2

Combine like terms: 40y2 + -40y2 = 0
128 + -38y2 = 0

Add '-128' to each side of the equation.
128 + -128 + -38y2 = 0 + -128

Combine like terms: 128 + -128 = 0
0 + -38y2 = 0 + -128
-38y2 = 0 + -128

Combine like terms: 0 + -128 = -128
-38y2 = -128

Divide each side by '-38'.
y2 = 3.368421053

Simplifying
y2 = 3.368421053

Take the square root of each side:
y = {-1.835325871, 1.835325871}