The problem is to solve the quadratic equation $x^2 + 6x - 13 = 0$. We need to find the values of $x$ that satisfy this equation.

1. Problem Description

The problem is to solve the quadratic equation $x^2 + 6x - 13 = 0$. We need to find the values of $x$ that satisfy this equation.

2. Solution Steps

We can use the quadratic formula to solve for $x$. The quadratic formula is given by:

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

In our equation, $x^2 + 6x - 13 = 0$, we have $a = 1$, $b = 6$, and $c = -13$. Plugging these values into the quadratic formula, we get:

$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-13)}}{2(1)}$

$x = \frac{-6 \pm \sqrt{36 + 52}}{2}$

$x = \frac{-6 \pm \sqrt{88}}{2}$

We can simplify the square root of 88:

$\sqrt{88} = \sqrt{4 \cdot 22} = \sqrt{4} \cdot \sqrt{22} = 2\sqrt{22}$

So, our equation becomes:

$x = \frac{-6 \pm 2\sqrt{22}}{2}$

We can divide both terms in the numerator by 2:

$x = -3 \pm \sqrt{22}$

The two solutions are $x = -3 + \sqrt{22}$ and $x = -3 - \sqrt{22}$.

3. Final Answer

$-3 + \sqrt{22}, -3 - \sqrt{22}$