(a) \(y = ax^{2} - bx - 12\) When a = 1, b = 2 and y = 3. \(3 = x^{2} - 2x - 12\) \(x^{2} - 2x - 12 - 3 = 0 \implies x^{2} - 2x - 15 = 0\) \(x^{2} - 5x + 3x - 15 = 0\) \((x - 5)(x + 3) = 0\) \(\text{x = 5 or -3}\) (b) \(\sqrt{x^{2} + 1} = \frac{5}{4}\) Squaring both sides, \(x^{2} + 1 = \frac{25}{16}\) \(x^{2} = \frac{25}{16} - 1 = \frac{9}{16}\) \(x = \sqrt{\frac{9}{16}} = \pm \frac{3}{4}\) The positive value of x = \(\frac{3}{4}\).