1. **State the problem:** We have a rectangle where the length is seven less than twice the width. The area is 15 square meters. We need to find the width $x$. 2. **Define variables:** Let the width be $x$. Then the length is $2x - 7$. 3. **Write the area formula:** Area $= \text{length} \times \text{width}$. 4. **Set up the equation:** $$x(2x - 7) = 15$$ 5. **Expand and simplify:** $$2x^2 - 7x = 15$$ 6. **Bring all terms to one side:** $$2x^2 - 7x - 15 = 0$$ 7. **Solve the quadratic equation using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-7$, and $c=-15$. 8. **Calculate the discriminant:** $$\Delta = (-7)^2 - 4 \times 2 \times (-15) = 49 + 120 = 169$$ 9. **Calculate the roots:** $$x = \frac{7 \pm \sqrt{169}}{4} = \frac{7 \pm 13}{4}$$ 10. **Find the two possible values:** - $$x = \frac{7 + 13}{4} = \frac{20}{4} = 5$$ - $$x = \frac{7 - 13}{4} = \frac{-6}{4} = -1.5$$ 11. **Interpret the results:** Width cannot be negative, so $x = 5$ meters. **Final answer:** The width is $\boxed{5}$ meters.