1. Stated problem: Given triangle $\triangle ABC$ with sides $|AB|=6$, $|BC|=8$, and angle $\angle ABC=120^\circ$, find the length of side $|AC|$. 2. Formula used: To find the length of a side opposite a known angle, we use the Law of Cosines: $$|AC|^2 = |AB|^2 + |BC|^2 - 2 \cdot |AB| \cdot |BC| \cdot \cos(\angle ABC)$$ 3. Substitute the known values: $$|AC|^2 = 6^2 + 8^2 - 2 \cdot 6 \cdot 8 \cdot \cos(120^\circ)$$ 4. Calculate the cosine of $120^\circ$: $$\cos(120^\circ) = -\frac{1}{2}$$ 5. Substitute $\cos(120^\circ)$: $$|AC|^2 = 36 + 64 - 2 \cdot 6 \cdot 8 \cdot \left(-\frac{1}{2}\right)$$ 6. Simplify the multiplication: $$|AC|^2 = 36 + 64 + 2 \cdot 6 \cdot 8 \cdot \frac{1}{2}$$ 7. Calculate the product: $$2 \cdot 6 \cdot 8 \cdot \frac{1}{2} = \cancel{2} \cdot 6 \cdot 8 \cdot \frac{1}{\cancel{2}} = 6 \cdot 8 = 48$$ 8. Sum all terms: $$|AC|^2 = 36 + 64 + 48 = 148$$ 9. Find $|AC|$ by taking the square root: $$|AC| = \sqrt{148} = \sqrt{4 \cdot 37} = 2\sqrt{37}$$ Final answer: The length of side $|AC|$ is $2\sqrt{37}$ units.