Q.Solve by factorization: $x^2 + 7x + 12 = 0$.

Find two numbers with product $12$ and sum $7$: these are $3$ and $4$.

$x^2+3x+4x+12=0 \Rightarrow x(x+3)+4(x+3)=0 \Rightarrow (x+3)(x+4)=0$, so $x=-3$ or $x=-4$.

We need $p,q$ with $p\times q=1\times12=12$ and $p+q=7$. Testing: $3\times4=12$ and $3+4=7$. So $p=3,q=4$.

Split: $x^2+3x+4x+12=0$.

Group: $(x^2+3x)+(4x+12)=0 \Rightarrow x(x+3)+4(x+3)=0$.

Factor: $(x+3)(x+4)=0$, giving $x=-3$ or $x=-4$.

Check: $(-3)^2+7(-3)+12=9-21+12=0$ ✓. $(-4)^2+7(-4)+12=16-28+12=0$ ✓.

Alternative — Sridharacharya's formula: $D=7^2-4(1)(12)=49-48=1$; $x=\dfrac{-7\pm1}{2}$, giving $x=-3$ or $x=-4$ — matches.

When both roots are negative (as here), students sometimes make sign errors writing the factors — since the sum $b=7$ is positive and the product $c=12$ is positive, BOTH roots must be negative (two negatives multiply to a positive and add to a negative of what's shown — check the general form's sign convention carefully before finalizing).