Solve for the roots of the equation:
[tex]\[ x^2 - 14x + 58 = 0 \][/tex]

A. [tex]\(\{3 + 7i, 3 - 7i\}\)[/tex]
B. [tex]\(\{2 + 9i, 2 - 9i\}\)[/tex]
C. [tex]\(\{2 + 19i, 2 - 19i\}\)[/tex]
D. [tex]\(\{7 + 3i, 7 - 3i\}\)[/tex]



Answer :

To solve the quadratic equation [tex]\(x^2 - 14x + 58 = 0\)[/tex], we'll use the quadratic formula, which states that for any quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], the solutions for [tex]\(x\)[/tex] are given by:

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

For our specific equation [tex]\(x^2 - 14x + 58 = 0\)[/tex]:

- [tex]\(a = 1\)[/tex]
- [tex]\(b = -14\)[/tex]
- [tex]\(c = 58\)[/tex]

First, calculate the discriminant, which is [tex]\(b^2 - 4ac\)[/tex]:

[tex]\[ b^2 - 4ac = (-14)^2 - 4 \cdot 1 \cdot 58 = 196 - 232 = -36 \][/tex]

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

Next, apply the quadratic formula:

[tex]\[ x = \frac{-(-14) \pm \sqrt{-36}}{2 \cdot 1} = \frac{14 \pm \sqrt{-36}}{2} = \frac{14 \pm 6i}{2} \][/tex]

This simplifies to:

[tex]\[ x = \frac{14}{2} \pm \frac{6i}{2} = 7 \pm 3i \][/tex]

Thus, the solutions are:

[tex]\[ x = 7 + 3i \quad \text{and} \quad x = 7 - 3i \][/tex]

Therefore, the correct answer is:

D. [tex]\(\{7+3i, 7-3i\}\)[/tex]