Q. \(x^2+3x-18=0\)

We want to solve the quadratic equation \(x^2 + 3x – 18 = 0\).

Step 1: Identify the method

Since this is a quadratic equation of the form \(ax^2 + bx + c = 0\) with \(a = 1\), \(b = 3\), and \(c = -18\), we can use factoring (if it factors nicely) or the quadratic formula. We’ll try factoring first.

Step 2: Find two numbers that multiply to \(ac\) and add to \(b\)

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

• \(mn = a \cdot c = 1 \cdot (-18) = -18\)
• \(m + n = b = 3\)

Now check factor pairs of \(-18\). The pair \((6, -3)\) works because:

• \(6 \cdot (-3) = -18\)
• \(6 + (-3) = 3\)

Step 3: Rewrite the quadratic using these factors

We rewrite \(x^2 + 3x – 18\) as:

\[
x^2 + 3x – 18 = (x + 6)(x – 3)
\]

So the equation becomes:

\[
(x + 6)(x – 3) = 0
\]

Step 4: Apply the zero-product property

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

So we set each factor equal to zero:

\[
x + 6 = 0
\]
\[
x – 3 = 0
\]

Step 5: Solve each simple equation

First equation:

\[
x + 6 = 0
\]

Subtract \(6\) from both sides:

\[
x = -6
\]

Second equation:

\[
x – 3 = 0
\]

Add \(3\) to both sides:

\[
x = 3
\]

Final Answer

The solutions to \(x^2 + 3x – 18 = 0\) are:

\[
x = -6 \quad \text{or} \quad x = 3
\]