Q. \(x^2 – 6x + 4 = 0\)

We need to solve the quadratic equation

\[
x^2 – 6x + 4 = 0
\]

Step 1: Identify the coefficients.

For a quadratic equation in the form \(ax^2 + bx + c = 0\), we have:

\[
a = 1,\quad b = -6,\quad c = 4
\]

Step 2: Use the quadratic formula.

The quadratic formula is

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

Step 3: Substitute \(a\), \(b\), and \(c\).

Substitute \(a = 1\), \(b = -6\), and \(c = 4\):

\[
x = \frac{-(-6) \pm \sqrt{(-6)^2 – 4(1)(4)}}{2(1)}
\]

Step 4: Simplify step-by-step.

First simplify the numerator outside the square root:

\[
-(-6) = 6
\]

Next compute the discriminant inside the square root:

\[
(-6)^2 – 4(1)(4) = 36 – 16 = 20
\]

So the formula becomes:

\[
x = \frac{6 \pm \sqrt{20}}{2}
\]

Step 5: Simplify the square root.

We simplify \(\sqrt{20}\):

\[
\sqrt{20} = \sqrt{4 \cdot 5} = \sqrt{4}\sqrt{5} = 2\sqrt{5}
\]

Substitute back:

\[
x = \frac{6 \pm 2\sqrt{5}}{2}
\]

Now divide every term in the numerator by \(2\):

\[
x = \frac{6}{2} \pm \frac{2\sqrt{5}}{2} = 3 \pm \sqrt{5}
\]

Final Answer:

\[
x = 3 + \sqrt{5}\quad \text{or}\quad x = 3 – \sqrt{5}
\]