1. **State the problem:** Solve the equation $$\frac{6-x}{2x} = \frac{3}{x+1}$$ and rewrite it as a quadratic equation. 2. **Use the cross-multiplication rule:** For two fractions equal to each other, $$\frac{a}{b} = \frac{c}{d}$$, cross-multiply to get $$a \cdot d = b \cdot c$$. 3. **Apply cross-multiplication:** $$ (6 - x)(x + 1) = 3 \cdot 2x $$ 4. **Expand both sides:** $$ (6 - x)(x + 1) = 6x $$ $$ 6x + 6 - x^2 - x = 6x $$ 5. **Simplify the left side:** $$ 6x + 6 - x^2 - x = 6x $$ $$ (6x - x) + 6 - x^2 = 6x $$ $$ 5x + 6 - x^2 = 6x $$ 6. **Bring all terms to one side to form a quadratic:** $$ 5x + 6 - x^2 - 6x = 0 $$ $$ -x^2 - x + 6 = 0 $$ 7. **Multiply both sides by -1 to get standard form:** $$ x^2 + x - 6 = 0 $$ **Final quadratic equation:** $$x^2 + x - 6 = 0$$