Question: Solve for $x$ using the quadratic formula.
$2x^2 + 6x - 36 = 0$
$x = [?], [ ]$
Enter the smallest solution first.
Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
1. **State the problem:** Solve the quadratic equation $$2x^2 + 6x - 36 = 0$$ using the quadratic formula.
2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions for $$x$$ are given by:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
3. **Identify coefficients:** Here, $$a = 2$$, $$b = 6$$, and $$c = -36$$.
4. **Calculate the discriminant:**
$$b^2 - 4ac = 6^2 - 4 \times 2 \times (-36) = 36 + 288 = 324$$
5. **Find the square root of the discriminant:**
$$\sqrt{324} = 18$$
6. **Apply the quadratic formula:**
$$x = \frac{-6 \pm 18}{2 \times 2} = \frac{-6 \pm 18}{4}$$
7. **Calculate each solution:**
- For the plus sign:
$$x = \frac{-6 + 18}{4} = \frac{12}{4} = 3$$
- For the minus sign:
$$x = \frac{-6 - 18}{4} = \frac{-24}{4} = -6$$
8. **Order the solutions from smallest to largest:**
$$x = -6, 3$$
**Final answer:** $$x = -6, 3$$