To solve the quadratic equation \(x^2 + 2x - 15 = 0\), we can use the quadratic formula, which is given by:

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

In this equation, \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\). For the given equation \(x^2 + 2x - 15 = 0\):

- \(a = 1\)
- \(b = 2\)
- \(c = -15\)

Now, we can plug these values into the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} \]

First, calculate the discriminant (\(b^2 - 4ac\)):

\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-15) = 4 + 60 = 64 \]

Now, substitute the discriminant back into the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{64}}{2} \]

Since \(\sqrt{64} = 8\), we have:

\[ x = \frac{-2 \pm 8}{2} \]

This gives us two solutions:

1. \( x = \frac{-2 + 8}{2} = \frac{6}{2} = 3 \)
2. \( x = \frac{-2 - 8}{2} = \frac{-10}{2} = -5 \)

So, the solutions to the equation \(x^2 + 2x - 15 = 0\) are:

\[ x = 3 \quad \text{and} \quad x = -5 \]