20(y^2+1)=4(y+5)

Solution for 20(y^2+1)=4(y+5) equation:

Simplifying
20(y2 + 1) = 4(y + 5)

Reorder the terms:
20(1 + y2) = 4(y + 5)
(1 * 20 + y2 * 20) = 4(y + 5)
(20 + 20y2) = 4(y + 5)

Reorder the terms:
20 + 20y2 = 4(5 + y)
20 + 20y2 = (5 * 4 + y * 4)
20 + 20y2 = (20 + 4y)

Add '-20' to each side of the equation.
20 + -20 + 20y2 = 20 + -20 + 4y

Combine like terms: 20 + -20 = 0
0 + 20y2 = 20 + -20 + 4y
20y2 = 20 + -20 + 4y

Combine like terms: 20 + -20 = 0
20y2 = 0 + 4y
20y2 = 4y

Solving
20y2 = 4y

Solving for variable 'y'.

Reorder the terms:
-4y + 20y2 = 4y + -4y

Combine like terms: 4y + -4y = 0
-4y + 20y2 = 0

Factor out the Greatest Common Factor (GCF), '4y'.
4y(-1 + 5y) = 0

Ignore the factor 4.
Subproblem 1
Set the factor 'y' equal to zero and attempt to solve:

Simplifying
y = 0

Solving
y = 0

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

Simplifying
y = 0
Subproblem 2
Set the factor '(-1 + 5y)' equal to zero and attempt to solve:

Simplifying
-1 + 5y = 0

Solving
-1 + 5y = 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 + 5y = 0 + 1

Combine like terms: -1 + 1 = 0
0 + 5y = 0 + 1
5y = 0 + 1

Combine like terms: 0 + 1 = 1
5y = 1

Divide each side by '5'.
y = 0.2

Simplifying
y = 0.2
Solution
y = {0, 0.2}