You have the quadratic equation:

$x^2 + 3x + 1 = 0$

We can solve this using the quadratic formula:

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

For your equation $x^2 + 3x + 1 = 0$, the coefficients are:

• $a = 1$
• $b = 3$
• $c = 1$

Now, substitute these values into the quadratic formula:

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

$x = \frac{-3 \pm \sqrt{9 - 4}}{2}$

$x = \frac{-3 \pm \sqrt{5}}{2}$

Thus, the two solutions are:

$x = \frac{-3 + \sqrt{5}}{2} \quad \text{or} \quad x = \frac{-3 - \sqrt{5}}{2}$

These are the exact solutions to the equation.

Would you like to explore more details or check something else about this equation?

Here are 5 related questions you might find interesting:

1. How do you complete the square to solve a quadratic equation?
2. How can you graph a quadratic equation like $x^2 + 3x + 1 = 0$?
3. What is the discriminant, and what does it tell you about the solutions?
4. How do you solve a quadratic equation by factoring?
5. What is the significance of the coefficients $a$, $b$, and $c$ in a quadratic equation?

Tip: The discriminant ($b^2 - 4ac$) tells you how many real solutions there are. If it's positive, there are two real solutions; if it's zero, there is exactly one real solution; if it's negative, the solutions are complex.