\begin{align} a & < 0 \phantom{000000} [\text{Condition #1}] \\ \\ \\ b^2 - 4ac & = (a)^2 - 4(a)(c) \\ & = a^2 - 4ac \\ \\ b^2 - 4ac & < 0 \\ a^2 - 4ac & < 0 \phantom{000000} [\text{Condition #2}] \\ \\ \text{Let } & a = -1, \phantom{000} [\text{to satisfy condition #1}] \\ (-1)^2 - 4(-1)c & < 0 \\ 1 + 4c & < 0 \\ 4c & < - 1 \\ c & < - {1 \over 4} \\ \\ \\ \therefore a & = -1, c = -1 \end{align}

\begin{align} \text{When } & C = 10, \\ 10 & = 1.2n^2 - 14.4n + 53.7 \\ 0 & = 1.2n^2 -14.4n + 43.7 \\ \\ [a = 1.2, b & = -14.4, c = 43.7] \\ \\ b^2 - 4ac & = (-14.4)^2 - 4(1.2)(43.7) \\ & = -2.4 < 0 \\ \\ \text{Since } b^2 - 4ac < 0, & \phantom{0} \text{equation has no real solutions}. \\ \\ \therefore \text{It is not possible to have } & \text{a cost of production of 10 thousand dollars.} \end{align}

\begin{align} p(x + 1) & = 3 - {1 \over 2}x^2 \\ px + p & = 3 - {1 \over 2}x^2 \\ {1 \over 2}x^2 + px + (p - 3) & = 0 \\ \\ b^2 - 4ac & = (p)^2 - 4 \left(1 \over 2\right)(p - 3) \\ & = p^2 - 2(p - 3) \\ & = p^2 - 2p + 6 \\ & = p^2 - 2p + \left(2 \over 2\right)^2 - \left(2 \over 2\right)^2 + 6 \phantom{000000} [\text{Complete the square}] \\ & = (p - 1)^2 - 1 + 6 \\ & = (p - 1)^2 + 5 \\ \\ \text{For all real val} & \text{ues of } p, \\ (p - 1)^2 & \ge 0 \\ (p - 1)^2 + 5 & \ge 5 \\ \\ \\ \therefore \text{For all real values of }p, \phantom{.} & b^2 - 4ac > 0 \text{ and equation has two real and distinct roots} \end{align}