To solve the quadratic equation \(x^2 + 4x + 2 = 0\), we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In the equation \(ax^2 + bx + c = 0\), we have:
- \(a = 1\)
- \(b = 4\)
- \(c = 2\)
Now we can plug these values into the quadratic formula.
First, we compute the discriminant (\(b^2 - 4ac\)):
\[ b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 2 = 16 - 8 = 8 \]
Since the discriminant is positive (\(8 > 0\)), there are two distinct real solutions.
Now substituting \(a\), \(b\), and the discriminant into the quadratic formula:
\[ x = \frac{-4 \pm \sqrt{8}}{2 \cdot 1} \]
Since \(\sqrt{8} = 2\sqrt{2}\), we can simplify further:
\[ x = \frac{-4 \pm 2\sqrt{2}}{2} \]
This simplifies to:
\[ x = -2 \pm \sqrt{2} \]
Thus, the two solutions are:
- \(x_1 = -2 + \sqrt{2}\)
- \(x_2 = -2 - \sqrt{2}\)
You can enter these values in the boxes provided. Here are the solutions:
- Exact solution 1: \(-2 + \sqrt{2}\)
- Exact solution 2: \(-2 - \sqrt{2}\)