Q. \(x^2-5x-14=0\)

We want to solve the quadratic equation

\[
x^{2}-5x-14=0
\]

Step 1: Factor the quadratic (if possible).

For a quadratic in the form \(ax^{2}+bx+c\), factoring works when we can write it as \((x-r)(x-s)=0\), where \(r\) and \(s\) satisfy:

\[
r+s = 5 \quad \text{and} \quad rs = -14
\]

We need two numbers whose product is \(-14\) and whose sum is \(5\).

Step 2: Find the two numbers.

The factor pairs of \(14\) are \(1\cdot 14\) and \(2\cdot 7\). To get a product of \(-14\), one number must be negative:

\[
(-1)\cdot 14 = -14 \quad \text{and} \quad (-2)\cdot 7 = -14
\]

Now check their sums:

\[
(-1)+14 = 13 \quad (\text{not } 5)
\]

\[
(-2)+7 = 5 \quad (\text{works})
\]

Step 3: Write the factored form.

Since \(r=-2\) and \(s=7\), we can factor the quadratic as:

\[
x^{2}-5x-14 = (x-7)(x+2)
\]

Step 4: Use the zero product property.

If \((x-7)(x+2)=0\), then either:

\[
x-7=0 \quad \text{or} \quad x+2=0
\]

Step 5: Solve each linear equation.

First equation:

\[
x-7=0
\]

Add \(7\) to both sides:

\[
x=7
\]

Second equation:

\[
x+2=0
\]

Subtract \(2\) from both sides:

\[
x=-2
\]

Final Answer:

\[
x=7 \quad \text{or} \quad x=-2
\]