12c^2-19c+5=0

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Solution for 12c^2-19c+5=0 equation:


Simplifying
12c2 + -19c + 5 = 0

Reorder the terms:
5 + -19c + 12c2 = 0

Solving
5 + -19c + 12c2 = 0

Solving for variable 'c'.

Factor a trinomial.
(1 + -3c)(5 + -4c) = 0

Subproblem 1

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

Subproblem 2

Set the factor '(5 + -4c)' equal to zero and attempt to solve: Simplifying 5 + -4c = 0 Solving 5 + -4c = 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 + -4c = 0 + -5 Combine like terms: 5 + -5 = 0 0 + -4c = 0 + -5 -4c = 0 + -5 Combine like terms: 0 + -5 = -5 -4c = -5 Divide each side by '-4'. c = 1.25 Simplifying c = 1.25

Solution

c = {0.3333333333, 1.25}

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