To solve the quadratic equation \( x^2 + 12x + 1 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 12 \), and \( c = 1 \).
Now, we can calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 12^2 - 4 \times 1 \times 1 = 144 - 4 = 140 \]
Since the discriminant is positive (140 > 0), there are two distinct real solutions.
Using the quadratic formula:
\[ x = \frac{-12 \pm \sqrt{140}}{2 \cdot 1} \]
Now we simplify \( \sqrt{140} \):
\[ \sqrt{140} = \sqrt{4 \times 35} = 2\sqrt{35} \]
Now substituting this back into the formula:
\[ x = \frac{-12 \pm 2\sqrt{35}}{2} \]
This simplifies to:
\[ x = -6 \pm \sqrt{35} \]
Hence, the two solutions are:
\[ x_1 = -6 + \sqrt{35} \quad \text{and} \quad x_2 = -6 - \sqrt{35} \]
You can write the exact solutions in the boxes as follows:
- First solution: \( -6 + \sqrt{35} \)
- Second solution: \( -6 - \sqrt{35} \)