Q. \(x^2+7x+12=0\)

We want to solve the equation

\[x^2+7x+12=0.\]

Step 1: Factor the quadratic.

We try to write the quadratic in the form

\[x^2+7x+12=(x+a)(x+b).\]

When you expand \((x+a)(x+b)\), you get

\[(x+a)(x+b)=x^2+(a+b)x+ab.\]

So we need numbers \(a\) and \(b\) such that:

\[a+b=7,\]

\[ab=12.\]

The factors of \(12\) are \(1\cdot 12\), \(2\cdot 6\), and \(3\cdot 4\). We check pairs that add to \(7\):

\[3+4=7,\quad 3\cdot 4=12.\]

So we can factor the quadratic as

\[x^2+7x+12=(x+3)(x+4).\]

Step 2: Use the zero-product property.

Substitute the factored form into the equation:

\[(x+3)(x+4)=0.\]

The zero-product property says that if the product of two expressions is zero, then at least one of the expressions must be zero. So we set:

\[x+3=0 \quad \text{or} \quad x+4=0.\]

Step 3: Solve each linear equation.

First:

\[x+3=0.\]

Subtract \(3\) from both sides:

\[x=-3.\]

Second:

\[x+4=0.\]

Subtract \(4\) from both sides:

\[x=-4.\]

Final answer:

\[x=-3 \quad \text{or} \quad x=-4.\]