Quadratic Solve 36Ecbb

1. **State the problem:** We need to solve the quadratic equation $$18abx^2 + (5a^2 - 7b^2)x - 9ab = 0$$ for $x$ using the quadratic formula. 2. **Recall the quadratic formula:** For a quadratic equation $$Ax^2 + Bx + C = 0$$, the solutions for $x$ are given by: $$x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$ 3. **Identify coefficients:** Here, $$A = 18ab$$ $$B = 5a^2 - 7b^2$$ $$C = -9ab$$ 4. **Calculate the discriminant:** $$\Delta = B^2 - 4AC = (5a^2 - 7b^2)^2 - 4 \times 18ab \times (-9ab)$$ First, expand the square: $$(5a^2 - 7b^2)^2 = (5a^2)^2 - 2 \times 5a^2 \times 7b^2 + (7b^2)^2 = 25a^4 - 70a^2b^2 + 49b^4$$ Calculate the product: $$-4 \times 18ab \times (-9ab) = 4 \times 18 \times 9 \times a^2 b^2 = 648 a^2 b^2$$ So, $$\Delta = 25a^4 - 70a^2b^2 + 49b^4 + 648 a^2 b^2 = 25a^4 + ( -70 + 648 ) a^2 b^2 + 49b^4 = 25a^4 + 578 a^2 b^2 + 49b^4$$ 5. **Write the solutions:** $$x = \frac{-(5a^2 - 7b^2) \pm \sqrt{25a^4 + 578 a^2 b^2 + 49b^4}}{2 \times 18ab} = \frac{-5a^2 + 7b^2 \pm \sqrt{25a^4 + 578 a^2 b^2 + 49b^4}}{36ab}$$ 6. **Summary:** The solutions for $x$ are: $$\boxed{x = \frac{-5a^2 + 7b^2 \pm \sqrt{25a^4 + 578 a^2 b^2 + 49b^4}}{36ab}}$$ This completes the solution using the quadratic formula.