5(k-2)+2[k-3(k+2)]=0

Solution for 5(k-2)+2[k-3(k+2)]=0 equation:

Simplifying
5(k + -2) + 2[k + -3(k + 2)] = 0

Reorder the terms:
5(-2 + k) + 2[k + -3(k + 2)] = 0
(-2 * 5 + k * 5) + 2[k + -3(k + 2)] = 0
(-10 + 5k) + 2[k + -3(k + 2)] = 0

Reorder the terms:
-10 + 5k + 2[k + -3(2 + k)] = 0
-10 + 5k + 2[k + (2 * -3 + k * -3)] = 0
-10 + 5k + 2[k + (-6 + -3k)] = 0

Reorder the terms:
-10 + 5k + 2[-6 + k + -3k] = 0

Combine like terms: k + -3k = -2k
-10 + 5k + 2[-6 + -2k] = 0
-10 + 5k + [-6 * 2 + -2k * 2] = 0
-10 + 5k + [-12 + -4k] = 0

Reorder the terms:
-10 + -12 + 5k + -4k = 0

Combine like terms: -10 + -12 = -22
-22 + 5k + -4k = 0

Combine like terms: 5k + -4k = 1k
-22 + 1k = 0

Solving
-22 + 1k = 0

Solving for variable 'k'.

Move all terms containing k to the left, all other terms to the right.

Add '22' to each side of the equation.
-22 + 22 + 1k = 0 + 22

Combine like terms: -22 + 22 = 0
0 + 1k = 0 + 22
1k = 0 + 22

Combine like terms: 0 + 22 = 22
1k = 22

Divide each side by '1'.
k = 22

Simplifying
k = 22