Absolute Value!

Solve for k over the real numbers:
abs(2-3 k) = 2 abs(k+2)

abs(2-3 k) = 2 abs(k+2) if and only if 2-3 k = 2 (k+2) or 2-3 k = -2 (k+2):
2-3 k = 2 (k+2) or 2-3 k = -2 (k+2)

Expand out terms of the right hand side:
2-3 k = 2 k+4 or 2-3 k = -2 (k+2)

Subtract 2 k+2 from both sides:
-5 k = 2 or 2-3 k = -2 (k+2)

Divide both sides by -5:
k = -2/5 or 2-3 k = -2 (k+2)

Expand out terms of the right hand side:
k = -2/5 or 2-3 k = -4-2 k

Subtract 2-2 k from both sides:
k = -2/5 or -k = -6

Multiply both sides by -1:
Answer: |k = -2/5     or     k = 6

|3k-2|=2|k+2|

Formula:

$\begin{array}{|rcll|} \hline |a|^2 &=& a^2 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline |3k-2| &=& 2|k+2| \qquad | \qquad \text{square both sides}\\ (3k-2)^2 &=& 2^2\cdot (k+2)^2 \\ 9k^2-12k+4 &=& 4\cdot (k^2+4k+4) \\ 9k^2-12k+4 &=& 4k^2+16k+16 \\ \dots \\ 5k^2-28k-12 &=& 0 \\\\ k_{1,2} &=& \frac{28 \pm \sqrt{28^2-4\cdot 5\cdot (-12) } }{2\cdot 5} \\ k_{1,2} &=& \frac{28 \pm \sqrt{1024 } }{10} \\ k_{1,2} &=& \frac{28 \pm 32 } {10} \\\\ k_1 &=& \frac{28 + 32 } {10} \\ k_1 &=& \frac{ 60 } {10} \\ \mathbf{ k_1 }& \mathbf{=} & \mathbf{6} \\\\ k_2 &=& \frac{28 - 32 } {10} \\ k_2 &=& \frac{ -4 } {10} \\ \mathbf{ k_2 } & \mathbf{=} & \mathbf{-0.4} \\ \hline \end{array}$