Solve for x in the equation x squared minus 12 x + 59 = 0.
x = negative 12 plus-or-minus StartRoot 85 EndRoot
x = negative 6 plus-or-minus StartRoot 23 EndRoot i
x = 6 plus-or-minus StartRoot 23 EndRoot i
x = 12 plus-or-minus StartRoot 85 EndRoot

The given equation is:

\[
x^2 - 12x + 59 = 0
\]

This is a quadratic equation. To solve it, we can use the quadratic formula:

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

For the equation \(x^2 - 12x + 59 = 0\), the coefficients are:
- \(a = 1\)
- \(b = -12\)
- \(c = 59\)

### Step 1: Calculate the discriminant \(\Delta\)
\[
\Delta = b^2 - 4ac = (-12)^2 - 4(1)(59) = 144 - 236 = -92
\]

Since the discriminant is negative, the solutions will be complex.

### Step 2: Apply the quadratic formula
\[
x = \frac{-(-12) \pm \sqrt{-92}}{2(1)} = \frac{12 \pm \sqrt{-92}}{2}
\]
\[
x = \frac{12 \pm \sqrt{92}i}{2}
\]

### Step 3: Simplify
\[
\sqrt{92} = \sqrt{4 \times 23} = 2\sqrt{23}
\]
\[
x = \frac{12 \pm 2\sqrt{23}i}{2}
\]
\[
x = 6 \pm \sqrt{23}i
\]

### Final Solution:
\[
x = 6 \pm \sqrt{23}i
\]

Thus, the correct solution is:

\[
x = 6 \pm \sqrt{23}i
\]

This matches the third option:

**\(x = 6 \pm \sqrt{23}i\)**

Would you like further details or have any questions?

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Here are 5 related questions to explore:
1. How do you interpret the discriminant in a quadratic equation?
2. What is the geometric meaning of complex roots?
3. How can you simplify expressions with square roots of negative numbers?
4. Can you apply the quadratic formula to a general quadratic equation with any coefficients?
5. How would this problem change if the discriminant were positive?

**Tip:** When the discriminant is negative, the roots of a quadratic equation are always complex and involve imaginary numbers.

Solve for x in the equation x squared minus 12 x + 59 = 0.

Learn how to solve the quadratic equation x^2 - 12x + 59 using the quadratic formula. This problem involves algebraic concepts, complex numbers, and discriminant analysis, with a detailed step-by-step solution. Ideal for high school students in Grades 10-12.

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