20c^2-21c+4=0

Solution for 20c^2-21c+4=0 equation:

Simplifying
20c2 + -21c + 4 = 0

Reorder the terms:
4 + -21c + 20c2 = 0

Solving
4 + -21c + 20c2 = 0

Solving for variable 'c'.

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

Subproblem 1
Set the factor '(1 + -4c)' equal to zero and attempt to solve:

Simplifying
1 + -4c = 0

Solving
1 + -4c = 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 + -4c = 0 + -1

Combine like terms: 1 + -1 = 0
0 + -4c = 0 + -1
-4c = 0 + -1

Combine like terms: 0 + -1 = -1
-4c = -1

Divide each side by '-4'.
c = 0.25

Simplifying
c = 0.25
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 = {0.25, 0.8}