Q. \(x^2 – 4x + 5 = 0\)

We want to solve the equation \(x^2 – 4x + 5 = 0\).

Step 1: Identify a, b, c.

This is a quadratic equation of the form \(ax^2 + bx + c = 0\).

Comparing with \(x^2 – 4x + 5 = 0\), we have:

• \(a = 1\)
• \(b = -4\)
• \(c = 5\)

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 \(c\) into the formula.

\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(5)}}{2(1)}
\]

Step 4: Simplify each part.

First, simplify \(-(-4)\):

\[
x = \frac{4 \pm \sqrt{(-4)^2 – 4(1)(5)}}{2}
\]

Next, compute \((-4)^2\) and \(4(1)(5)\):

\[
(-4)^2 = 16
\]
\[
4(1)(5) = 20
\]

Substitute these back in:

\[
x = \frac{4 \pm \sqrt{16 – 20}}{2}
\]

Now compute the number inside the square root:

\[
16 – 20 = -4
\]

So:

\[
x = \frac{4 \pm \sqrt{-4}}{2}
\]

Step 5: Simplify \(\sqrt{-4}\).

Write \(-4\) as \(-1 \cdot 4\):

\[
\sqrt{-4} = \sqrt{-1 \cdot 4} = \sqrt{-1}\sqrt{4} = i \cdot 2 = 2i
\]

So the equation becomes:

\[
x = \frac{4 \pm 2i}{2}
\]

Step 6: Divide each term in the numerator by \(2\).

\[
x = \frac{4}{2} \pm \frac{2i}{2} = 2 \pm i
\]

Step 7: State the solutions.

The two solutions are:

\[
x = 2 + i
\]
\[
x = 2 – i
\]

Final Answer: \(\displaystyle x = 2 \pm i\).