1. Paper 2 Problem: Solve the quadratic equation $2x^2 - 5x + 3 = 0$. Step 1: Identify coefficients: $a=2$, $b=-5$, $c=3$. Step 2: Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Step 3: Calculate the discriminant: $$\Delta = (-5)^2 - 4 \times 2 \times 3 = 25 - 24 = 1$$. Step 4: Substitute values: $$x = \frac{5 \pm \sqrt{1}}{4} = \frac{5 \pm 1}{4}$$. Step 5: Find roots: $$x_1 = \frac{5 + 1}{4} = \frac{6}{4} = 1.5$$, $$x_2 = \frac{5 - 1}{4} = \frac{4}{4} = 1$$. Answer: $x = 1.5$ or $x = 1$. 2. Paper 4 Problem: A circle has equation $x^2 + y^2 - 6x + 8y + 9 = 0$. Find its center and radius. Step 1: Rewrite the equation grouping $x$ and $y$ terms: $$x^2 - 6x + y^2 + 8y = -9$$. Step 2: Complete the square for $x$: $$x^2 - 6x = (x - 3)^2 - 9$$. Step 3: Complete the square for $y$: $$y^2 + 8y = (y + 4)^2 - 16$$. Step 4: Substitute back: $$(x - 3)^2 - 9 + (y + 4)^2 - 16 = -9$$. Step 5: Simplify: $$(x - 3)^2 + (y + 4)^2 = -9 + 9 + 16 = 16$$. Answer: Center is $(3, -4)$ and radius is $\sqrt{16} = 4$.