Question: 15. Find the solution for the quadratic inequality using the quadratic formula. Use $-3$, $0$, $4$ as test values.
$2x^2 - 4x - 6 < 0$
$1 > x > -3$
$-1 < x < 3$
$x < -1$ or $x > 3$
$-3 < x < 1$
1. **State the problem:** Solve the quadratic inequality $$2x^2 - 4x - 6 < 0$$ using the quadratic formula and test values $-3$, $0$, and $4$.
2. **Recall the quadratic formula:** For a quadratic equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$.
3. **Identify coefficients:** Here, $a = 2$, $b = -4$, and $c = -6$.
4. **Calculate the discriminant:**
$$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64$$
5. **Find the roots:**
$$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}$$
6. **Calculate each root:**
- $$x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3$$
- $$x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1$$
7. **Determine intervals for inequality:** Since $a = 2 > 0$, the parabola opens upward, so the quadratic expression is less than zero between the roots.
8. **Solution interval:**
$$-1 < x < 3$$
9. **Test values:**
- At $x = -3$: $$2(-3)^2 - 4(-3) - 6 = 18 + 12 - 6 = 24 > 0$$ (False for inequality)
- At $x = 0$: $$2(0)^2 - 4(0) - 6 = -6 < 0$$ (True for inequality)
- At $x = 4$: $$2(4)^2 - 4(4) - 6 = 32 - 16 - 6 = 10 > 0$$ (False for inequality)
10. **Final answer:** The solution to the inequality is $$\boxed{-1 < x < 3}$$.