1. **State the problem:** Solve the equation $$2x(3x + 17) = 12$$ and check the solutions. 2. **Expand the left side:** Use the distributive property $$a(b+c) = ab + ac$$. $$2x \times 3x + 2x \times 17 = 12$$ $$6x^2 + 34x = 12$$ 3. **Bring all terms to one side to set the equation to zero:** $$6x^2 + 34x - 12 = 0$$ 4. **Simplify the quadratic equation by dividing all terms by 2:** $$\cancel{2} \times 3x^2 + \cancel{2} \times 17x - \cancel{2} \times 6 = 0$$ $$3x^2 + 17x - 6 = 0$$ 5. **Use the quadratic formula to solve for $x$:** The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=17$, and $c=-6$. 6. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 17^2 - 4 \times 3 \times (-6) = 289 + 72 = 361$$ 7. **Calculate the square root of the discriminant:** $$\sqrt{361} = 19$$ 8. **Find the two solutions:** $$x = \frac{-17 \pm 19}{2 \times 3} = \frac{-17 \pm 19}{6}$$ 9. **Calculate each root:** - For the plus sign: $$x = \frac{-17 + 19}{6} = \frac{2}{6} = \frac{1}{3}$$ - For the minus sign: $$x = \frac{-17 - 19}{6} = \frac{-36}{6} = -6$$ 10. **Check the solutions by substituting back into the original equation:** - For $x=\frac{1}{3}$: $$2 \times \frac{1}{3} \times (3 \times \frac{1}{3} + 17) = 2 \times \frac{1}{3} \times (1 + 17) = 2 \times \frac{1}{3} \times 18 = 12$$ - For $x=-6$: $$2 \times (-6) \times (3 \times (-6) + 17) = -12 \times (-18 + 17) = -12 \times (-1) = 12$$ Both solutions satisfy the equation. **Final answer:** The solution set is $$\left\{ \frac{1}{3}, -6 \right\}$$.