Q. \(x^2 – x – 20 = 0\)

We want to solve the equation

\[
x^2 – x – 20 = 0.
\]

This is a quadratic equation of the form \(ax^2+bx+c=0\), where \(a=1\), \(b=-1\), and \(c=-20\).

Step 1: Factor the quadratic

We want to rewrite \(x^2 – x – 20\) as a product of two binomials.

Since the first term is \(x^2\), the two binomials must look like \((x + \text{something})(x + \text{something})\).

Step 2: Find numbers that multiply to \(-20\) and add to \(-1\)

The product of the two constant terms must be \(-20\), and their sum must be \(-1\).

Let the constants be \(m\) and \(n\). Then we need:

\[
mn = -20
\]
\[
m+n = -1.
\]

Check factor pairs of \(-20\):

\[
-5 \cdot 4 = -20 \quad \text{and} \quad -5 + 4 = -1.
\]

So we can factor using \(-5\) and \(4\).

Step 3: Write the factored form

Now rewrite the quadratic:

\[
x^2 – x – 20 = (x-5)(x+4).
\]

So the equation becomes:

\[
(x-5)(x+4)=0.
\]

Step 4: Use the zero product property

The zero product property says: if \(AB=0\), then \(A=0\) or \(B=0\).

So either:

\[
x-5=0
\]

or

\[
x+4=0.
\]

Step 5: Solve each equation

First equation:

\[
x-5=0
\]
\[
x=5.
\]

Second equation:

\[
x+4=0
\]
\[
x=-4.
\]

Final Answer

The solutions to \(x^2-x-20=0\) are:

\[
x=5 \quad \text{or} \quad x=-4.
\]