To solve the quadratic equation:

$x^2 + 3x + 18 = 0$

we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where:

• $a = 1$
• $b = 3$
• $c = 18$

Step 1: Substitute the values

$x = \frac{-3 \pm \sqrt{(3)^2 - 4(1)(18)}}{2(1)}$

$x = \frac{-3 \pm \sqrt{9 - 72}}{2}$

Step 2: Simplify under the square root

$x = \frac{-3 \pm \sqrt{-63}}{2}$

Since the expression under the square root is negative ($-63$), the solutions will involve imaginary numbers.

Step 3: Express in terms of imaginary numbers

$x = \frac{-3 \pm i\sqrt{63}}{2}$

$x = \frac{-3 \pm i\sqrt{9 \cdot 7}}{2}$

$x = \frac{-3 \pm 3i\sqrt{7}}{2}$

Step 4: Write the final solution

$x = \frac{-3}{2} \pm \frac{3i\sqrt{7}}{2}$

So, the solutions are: $x = -\frac{3}{2} + \frac{3i\sqrt{7}}{2} \quad \text{and} \quad x = -\frac{3}{2} - \frac{3i\sqrt{7}}{2}$

Would you like a more detailed explanation, or do you have any questions?

Here are 5 related questions you might find interesting:

1. How can you determine if a quadratic equation has real or complex solutions?
2. What are the implications of having a negative discriminant ($b^2 - 4ac < 0$) in the quadratic formula?
3. Can we solve this equation by completing the square? How?
4. What does the graph of this quadratic equation look like on the complex plane?
5. How do imaginary numbers apply to real-world scenarios?

Tip: When the discriminant ($b^2 - 4ac$) is negative, the solutions of a quadratic equation will always be complex, involving imaginary numbers.