Q. \(x^2 + 2x + 5 = 0\)

We want to solve the equation

\[
x^2 + 2x + 5 = 0.
\]

Step 1: Identify the coefficients.

Compare with the standard quadratic form

\[
ax^2 + bx + c = 0.
\]

Here:

\[
a = 1,\quad b = 2,\quad c = 5.
\]

Step 2: Use the quadratic formula.

The solutions to \(ax^2 + bx + c = 0\) are given by

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

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

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

Step 4: Simplify inside the square root (the discriminant).

Compute \(b^2 – 4ac\):

\[
2^2 – 4(1)(5) = 4 – 20 = -16.
\]

Step 5: Substitute the discriminant back in.

\[
x = \frac{-2 \pm \sqrt{-16}}{2}.
\]

Step 6: Simplify the square root of a negative number.

Since \(-16 = 16(-1)\), we have

\[
\sqrt{-16} = \sqrt{16}\sqrt{-1} = 4i,
\]

where \(i\) is the imaginary unit with \(i^2 = -1\).

Step 7: Final simplification.

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

Split the division:

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

\[
x = -1 \pm 2i.
\]

So the solutions are:

\[
\boxed{x = -1 + 2i \quad \text{and} \quad x = -1 – 2i.}
\]