To solve the equation (x^2 - 5x + 6 = 0), we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}). Plugging in the values of (a = 1, b = -5, c = 6), we get (x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2} = 2, 3).

To find the area of a triangle with sides of length 6 cm, 8 cm, and 10 cm, we can use Heron's formula: (A = \sqrt{s(s-a)(s-b)(s-c)}), where (s) is the semiperimeter of the triangle and (a, b, c) are the lengths of the sides. Plugging in the values of (a = 6, b = 8, c = 10), we get (s = \frac{6 + 8 + 10}{2} = 12). Therefore, (A = \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12(6)(4)(2)} = 36) cm^2.

To find the greatest common divisor (GCD) of 12 and 18, we can use the Euclidean algorithm. We divide 18 by 12 to get a quotient of 1 and a remainder of 6. Then we divide 12 by 6 to get a quotient of 2 and a remainder of 0. Therefore, the GCD of 12 and 18 is 6.

To find the sum of the first 100 positive integers, we can use the formula (S = \frac{n(n+1)}{2}), where (n) is the number of integers. Plugging in (n = 100), we get (S = \frac{100(100+1)}{2} = \frac{100(101)}{2} = 5050).

To find the time at which the two trains will meet, we can use the formula (t = \frac{d}{s_1 + s_2}), where (t) is the time, (d) is the distance, and (s_1) and (s_2) are the speeds of the two trains. Plugging in the values of (d = 1200) km, (s_1 = 60) km/h, and (s_2 = 70) km/h, we get (t = \frac{1200}{60 + 70} = \frac{1200}{130} = 9.23) hours. Since the first train leaves at 10:00 AM and the second train leaves at 11:00 AM, the two trains will meet at 10:00 AM + 9.23 hours = 7:18 PM.