Q. \(x^2 – 3x – 10 = 0\)

We want to solve the equation

\[x^2 – 3x – 10 = 0.\]

Step 1: Identify the quadratic form

A quadratic equation has the form

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

Here, we can match coefficients:

\[a = 1,\quad b = -3,\quad c = -10.\]

Step 2: Factor the quadratic

The goal is to rewrite \(x^2 – 3x – 10\) as a product of two binomials:

\[(x + m)(x + n) = x^2 + (m+n)x + mn.\]

We need \(m+n = -3\) and \(mn = -10\).

Now list factor pairs of \(-10\):

• \(1 \cdot -10 = -10\)
• \(-1 \cdot 10 = -10\)
• \(2 \cdot -5 = -10\)
• \(-2 \cdot 5 = -10\)

Check which pair adds to \(-3\):

• \(2 + (-5) = -3\) ✓

So \(m = 2\) and \(n = -5\). Therefore,

\[x^2 – 3x – 10 = (x + 2)(x – 5).\]

Step 3: Use the zero product property

If a product is zero, then at least one factor must be zero. So:

\[(x + 2)(x – 5) = 0.\]

This gives two equations:

\[x + 2 = 0 \quad \text{or} \quad x – 5 = 0.\]

Step 4: Solve each linear equation

First:

\[x + 2 = 0.\]

Subtract \(2\) from both sides:

\[x = -2.\]

Second:

\[x – 5 = 0.\]

Add \(5\) to both sides:

\[x = 5.\]

Final Answer

The solutions to the equation \(x^2 – 3x – 10 = 0\) are:

\[x = -2 \quad \text{or} \quad x = 5.\]