(k+4)(3k-12)=0

Solution for (k+4)(3k-12)=0 equation:

Simplifying
(k + 4)(3k + -12) = 0

Reorder the terms:
(4 + k)(3k + -12) = 0

Reorder the terms:
(4 + k)(-12 + 3k) = 0

Multiply (4 + k) * (-12 + 3k)
(4(-12 + 3k) + k(-12 + 3k)) = 0
((-12 * 4 + 3k * 4) + k(-12 + 3k)) = 0
((-48 + 12k) + k(-12 + 3k)) = 0
(-48 + 12k + (-12 * k + 3k * k)) = 0
(-48 + 12k + (-12k + 3k2)) = 0

Combine like terms: 12k + -12k = 0
(-48 + 0 + 3k2) = 0
(-48 + 3k2) = 0

Solving
-48 + 3k2 = 0

Solving for variable 'k'.

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

Add '48' to each side of the equation.
-48 + 48 + 3k2 = 0 + 48

Combine like terms: -48 + 48 = 0
0 + 3k2 = 0 + 48
3k2 = 0 + 48

Combine like terms: 0 + 48 = 48
3k2 = 48

Divide each side by '3'.
k2 = 16

Simplifying
k2 = 16

Take the square root of each side:
k = {-4, 4}