Solve Quadratic 3Ccb41

1. **State the problem:** Solve the equation $$\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$$ using the quadratic equation method. 2. **Find a common denominator and combine the fractions on the left side:** $$\frac{(x-7) - (x+4)}{(x+4)(x-7)} = \frac{11}{30}$$ 3. **Simplify the numerator:** $$\frac{x - 7 - x - 4}{(x+4)(x-7)} = \frac{11}{30}$$ $$\frac{-11}{(x+4)(x-7)} = \frac{11}{30}$$ 4. **Cross multiply to clear denominators:** $$-11 \times 30 = 11 \times (x+4)(x-7)$$ $$-330 = 11(x+4)(x-7)$$ 5. **Divide both sides by 11:** $$\cancel{11} \times -30 = \cancel{11} (x+4)(x-7)$$ $$-30 = (x+4)(x-7)$$ 6. **Expand the right side:** $$-30 = x^2 - 7x + 4x - 28$$ $$-30 = x^2 - 3x - 28$$ 7. **Bring all terms to one side to set equation to zero:** $$x^2 - 3x - 28 + 30 = 0$$ $$x^2 - 3x + 2 = 0$$ 8. **Use the quadratic formula:** The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-3$$, and $$c=2$$. 9. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-3)^2 - 4(1)(2) = 9 - 8 = 1$$ 10. **Find the roots:** $$x = \frac{-(-3) \pm \sqrt{1}}{2(1)} = \frac{3 \pm 1}{2}$$ 11. **Calculate each root:** - $$x = \frac{3 + 1}{2} = \frac{4}{2} = 2$$ - $$x = \frac{3 - 1}{2} = \frac{2}{2} = 1$$ 12. **Check for restrictions:** The original denominators are $$x+4$$ and $$x-7$$, so $$x \neq -4$$ and $$x \neq 7$$. Both solutions $$x=1$$ and $$x=2$$ are valid. **Final answer:** $$x = 1$$ or $$x = 2$$