Q. \(x^{2}+2x-63=0\)

We want to solve the quadratic equation:

\[
x^2 + 2x – 63 = 0
\]

Step 1: Factor the quadratic.

A quadratic of the form \(ax^2 + bx + c\) can be factored if we find two numbers whose product is \(ac\) and whose sum is \(b\).

Here, \(a = 1\), \(b = 2\), and \(c = -63\).

So we need two numbers \(m\) and \(n\) such that:

\[
m \cdot n = -63
\]
\[
m + n = 2
\]

Step 2: Find the numbers.

Because the product is \(-63\), one number must be positive and the other negative.

Check factor pairs of \(63\): \(1\) and \(63\), \(3\) and \(21\).

We test which pair gives a sum of \(2\):

\[
(-1) + 3 = 2
\]

So we take \(m = 3\) and \(n = -1\). Their product is:

\[
3 \cdot (-1) = -3
\]

That is not \(-63\), so we must use the factor pair method more carefully: we need factors that multiply to \(-63\) and add to \(2\).

Try these factor pairs of \(-63\):

\(63\) and \(-1\):

\[
63 + (-1) = 62 \quad \text{not } 2
\]

\(21\) and \(-3\):

\[
21 + (-3) = 18 \quad \text{not } 2
\]

\(9\) and \(-7\):

\[
9 + (-7) = 2
\]

Great! So the two numbers are \(9\) and \(-7\).

Step 3: Rewrite and factor the quadratic.

Now factor the expression:

\[
x^2 + 2x – 63 = (x+9)(x-7)
\]

Step 4: Set each factor equal to zero.

Using the zero product property:

\[
(x+9)(x-7)=0
\]
\[
x+9=0 \quad \text{or} \quad x-7=0
\]

Step 5: Solve each equation.

First equation:

\[
x+9=0
\]
\[
x=-9
\]

Second equation:

\[
x-7=0
\]
\[
x=7
\]

Final Answer:

The solutions to \(x^2 + 2x – 63 = 0\) are:

\[
x = -9 \quad \text{and} \quad x = 7
\]