1. **State the problem:** We need to find all values of $x$ such that $k(x) = 3x^2 - 19x - 14 > 0$. 2. **Recall the formula and rules:** For a quadratic function $ax^2 + bx + c$, the sign of the function depends on the roots and the leading coefficient $a$. 3. **Find the roots of $k(x)$ by solving $3x^2 - 19x - 14 = 0$ using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=-19$, and $c=-14$. 4. **Calculate the discriminant:** $$\Delta = (-19)^2 - 4 \times 3 \times (-14) = 361 + 168 = 529$$ 5. **Calculate the roots:** $$x = \frac{19 \pm \sqrt{529}}{2 \times 3} = \frac{19 \pm 23}{6}$$ 6. **Find each root:** - For $+$ sign: $$x = \frac{19 + 23}{6} = \frac{42}{6} = 7$$ - For $-$ sign: $$x = \frac{19 - 23}{6} = \frac{-4}{6} = -\frac{2}{3}$$ 7. **Determine intervals where $k(x) > 0$:** Since $a=3 > 0$, the parabola opens upward, so $k(x) > 0$ outside the roots. 8. **Final solution:** $$x < -\frac{2}{3} \quad \text{or} \quad x > 7$$