To find the roots of the quadratic equation \(x^2 + 12x - 36 = 0\), we can use the quadratic formula:

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

Here, the coefficients of the equation \(ax^2 + bx + c = 0\) are:

• \(a = 1\)
• \(b = 12\)
• \(c = -36\)

Substitute these values into the quadratic formula:

\(x = \frac{-12 \pm \sqrt{(12)^2 - 4 \cdot 1 \cdot (-36)}}{2 \cdot 1}\)

\(x = \frac{-12 \pm \sqrt{144 + 144}}{2}\)

\(x = \frac{-12 \pm \sqrt{288}}{2}\)

Simplify \(\sqrt{288}\):

\(\sqrt{288} = \sqrt{144 \cdot 2} = 12\sqrt{2}\)

Now, substitute back:

\(x = \frac{-12 \pm 12\sqrt{2}}{2}\)

Divide each term by 2:

\(x = \frac{-12}{2} \pm \frac{12\sqrt{2}}{2}\)

\(x = -6 \pm 6\sqrt{2}\)

Therefore, the roots are:

• \(x_1 = -6 + 6\sqrt{2}\)
• \(x_2 = -6 - 6\sqrt{2}\)

Upon inspection, we can factor \(x_1\) and \(x_2\) with \(6(\sqrt{2} - 1)\):

• \(x_1 = -6 + 6\sqrt{2} = 6(\sqrt{2} - 1)\)
• \(x_2 = -6 - 6\sqrt{2} = -6(1 + \sqrt{2})\)

Thus, the roots of the equation \(x^2 + 12x - 36 = 0\) are

$-6(1 + \sqrt{2}), 6(\sqrt{2} - 1)$

, which matches the correct answer. Hence, the option

$-6(1 + \sqrt{2}), 6(\sqrt{2} - 1)$ is correct.