(c^2)+(2c)=35

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Solution for (c^2)+(2c)=35 equation:


Simplifying
(c2) + (2c) = 35
c2 + (2c) = 35

Reorder the terms:
(2c) + c2 = 35

Solving
(2c) + c2 = 35

Solving for variable 'c'.

Reorder the terms:
-35 + (2c) + c2 = 35 + -35

Combine like terms: 35 + -35 = 0
-35 + (2c) + c2 = 0

Factor a trinomial.
(-7 + -1c)(5 + -1c) = 0

Subproblem 1

Set the factor '(-7 + -1c)' equal to zero and attempt to solve: Simplifying -7 + -1c = 0 Solving -7 + -1c = 0 Move all terms containing c to the left, all other terms to the right. Add '7' to each side of the equation. -7 + 7 + -1c = 0 + 7 Combine like terms: -7 + 7 = 0 0 + -1c = 0 + 7 -1c = 0 + 7 Combine like terms: 0 + 7 = 7 -1c = 7 Divide each side by '-1'. c = -7 Simplifying c = -7

Subproblem 2

Set the factor '(5 + -1c)' equal to zero and attempt to solve: Simplifying 5 + -1c = 0 Solving 5 + -1c = 0 Move all terms containing c to the left, all other terms to the right. Add '-5' to each side of the equation. 5 + -5 + -1c = 0 + -5 Combine like terms: 5 + -5 = 0 0 + -1c = 0 + -5 -1c = 0 + -5 Combine like terms: 0 + -5 = -5 -1c = -5 Divide each side by '-1'. c = 5 Simplifying c = 5

Solution

c = {-7, 5}

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