20c^2-31c+12=0

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Solution for 20c^2-31c+12=0 equation:


Simplifying
20c2 + -31c + 12 = 0

Reorder the terms:
12 + -31c + 20c2 = 0

Solving
12 + -31c + 20c2 = 0

Solving for variable 'c'.

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

Subproblem 1

Set the factor '(3 + -4c)' equal to zero and attempt to solve: Simplifying 3 + -4c = 0 Solving 3 + -4c = 0 Move all terms containing c to the left, all other terms to the right. Add '-3' to each side of the equation. 3 + -3 + -4c = 0 + -3 Combine like terms: 3 + -3 = 0 0 + -4c = 0 + -3 -4c = 0 + -3 Combine like terms: 0 + -3 = -3 -4c = -3 Divide each side by '-4'. c = 0.75 Simplifying c = 0.75

Subproblem 2

Set the factor '(4 + -5c)' equal to zero and attempt to solve: Simplifying 4 + -5c = 0 Solving 4 + -5c = 0 Move all terms containing c to the left, all other terms to the right. Add '-4' to each side of the equation. 4 + -4 + -5c = 0 + -4 Combine like terms: 4 + -4 = 0 0 + -5c = 0 + -4 -5c = 0 + -4 Combine like terms: 0 + -4 = -4 -5c = -4 Divide each side by '-5'. c = 0.8 Simplifying c = 0.8

Solution

c = {0.75, 0.8}

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