1. The problem is to solve a more complex algebraic expression or equation. Since the user did not specify the exact problem, let's consider a typical example: Solve for $x$ in the equation $$2x^2 - 5x + 3 = 0$$. 2. The formula used to solve quadratic equations is the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. 3. Identify the coefficients: $a=2$, $b=-5$, $c=3$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-5)^2 - 4 \times 2 \times 3 = 25 - 24 = 1$$ 5. Since the discriminant is positive, there are two real solutions. 6. Substitute values into the quadratic formula: $$x = \frac{-(-5) \pm \sqrt{1}}{2 \times 2} = \frac{5 \pm 1}{4}$$ 7. Calculate each solution: - For the plus sign: $$x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{\cancel{6}}{\cancel{4}} = \frac{3}{2}$$ - For the minus sign: $$x = \frac{5 - 1}{4} = \frac{4}{4} = 1$$ 8. Final answer: $$x = \frac{3}{2} \text{ or } x = 1$$ This method can be applied to any quadratic equation to find its roots.