Question

Use the discriminant to determine how many real number solutions exist for the quadratic equation –4j^2 + 3j – 28 = 0

Answer 1

$\bf \begin{array}{lcccll} & -4j^2& +3j& -28\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \qquad discriminant\implies b^2-4ac \\\\\\ (3)^2-4(-4)(-28)= \begin{cases} 0&\textit{one solution}\\ positive&\textit{two solutions}\\ negative&\textit{no solution} \end{cases}$