2002 AMC 10A Problem 10

Both products on the left start with $(2x+3)$. That shared piece can be pulled out, just like $5 \times 4 + 5 \times 6 = 5 \times (4 + 6)$.

1. Pull out the common factor:
$$(2x+3)(x-4) + (2x+3)(x-6) = (2x+3)\big[(x-4) + (x-6)\big] = (2x+3)(2x-10).$$

2. Set each factor to zero. A product is zero exactly when one of its factors is zero:
- $2x + 3 = 0$ gives $x = -\frac{3}{2}$,
- $2x - 10 = 0$ gives $x = 5$.

3. Add the roots:
$$-\frac{3}{2} + 5 = \frac{7}{2}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{7}{2}}$.

When every term of an equation shares a factor, pulling it out is faster and safer than multiplying everything out, and it shows the roots immediately. Never divide both sides by an expression that contains $x$: that throws away the root that makes the expression zero.

Multiply out: $(2x+3)(x-4) = 2x^2 - 5x - 12$ and $(2x+3)(x-6) = 2x^2 - 9x - 18$. Adding gives $4x^2 - 14x - 30 = 0$. For a quadratic $ax^2 + bx + c = 0$, the roots add to $-\frac{b}{a}$, so here the sum is $\frac{14}{4} = \frac{7}{2}$.