25c^2+4+20c=c(c-15)-5

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Solution for 25c^2+4+20c=c(c-15)-5 equation:


Simplifying
25c2 + 4 + 20c = c(c + -15) + -5

Reorder the terms:
4 + 20c + 25c2 = c(c + -15) + -5

Reorder the terms:
4 + 20c + 25c2 = c(-15 + c) + -5
4 + 20c + 25c2 = (-15 * c + c * c) + -5
4 + 20c + 25c2 = (-15c + c2) + -5

Reorder the terms:
4 + 20c + 25c2 = -5 + -15c + c2

Solving
4 + 20c + 25c2 = -5 + -15c + c2

Solving for variable 'c'.

Reorder the terms:
4 + 5 + 20c + 15c + 25c2 + -1c2 = -5 + -15c + c2 + 5 + 15c + -1c2

Combine like terms: 4 + 5 = 9
9 + 20c + 15c + 25c2 + -1c2 = -5 + -15c + c2 + 5 + 15c + -1c2

Combine like terms: 20c + 15c = 35c
9 + 35c + 25c2 + -1c2 = -5 + -15c + c2 + 5 + 15c + -1c2

Combine like terms: 25c2 + -1c2 = 24c2
9 + 35c + 24c2 = -5 + -15c + c2 + 5 + 15c + -1c2

Reorder the terms:
9 + 35c + 24c2 = -5 + 5 + -15c + 15c + c2 + -1c2

Combine like terms: -5 + 5 = 0
9 + 35c + 24c2 = 0 + -15c + 15c + c2 + -1c2
9 + 35c + 24c2 = -15c + 15c + c2 + -1c2

Combine like terms: -15c + 15c = 0
9 + 35c + 24c2 = 0 + c2 + -1c2
9 + 35c + 24c2 = c2 + -1c2

Combine like terms: c2 + -1c2 = 0
9 + 35c + 24c2 = 0

Factor a trinomial.
(9 + 8c)(1 + 3c) = 0

Subproblem 1

Set the factor '(9 + 8c)' equal to zero and attempt to solve: Simplifying 9 + 8c = 0 Solving 9 + 8c = 0 Move all terms containing c to the left, all other terms to the right. Add '-9' to each side of the equation. 9 + -9 + 8c = 0 + -9 Combine like terms: 9 + -9 = 0 0 + 8c = 0 + -9 8c = 0 + -9 Combine like terms: 0 + -9 = -9 8c = -9 Divide each side by '8'. c = -1.125 Simplifying c = -1.125

Subproblem 2

Set the factor '(1 + 3c)' equal to zero and attempt to solve: Simplifying 1 + 3c = 0 Solving 1 + 3c = 0 Move all terms containing c to the left, all other terms to the right. Add '-1' to each side of the equation. 1 + -1 + 3c = 0 + -1 Combine like terms: 1 + -1 = 0 0 + 3c = 0 + -1 3c = 0 + -1 Combine like terms: 0 + -1 = -1 3c = -1 Divide each side by '3'. c = -0.3333333333 Simplifying c = -0.3333333333

Solution

c = {-1.125, -0.3333333333}

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