30)
[tex]\bf \begin{array}{llll}
y=&{{ a}}x^2&{{ +b}}x&{{ +c}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\\\\\\
discriminant\implies b^2-4ac=
\begin{cases}
0&\textit{one solution}\\
positive&\textit{two solutions}\\
negative&\textit{no solution}
\end{cases}[/tex]
31)
a simple case for that would just be, using an equation with an imaginary value, let's do so
[tex]\bf \sqrt{-5}=x\implies \sqrt{-1\cdot 5}=x\implies \sqrt{-1}\sqrt{5}=x\implies i\sqrt{5}=x\\\\
-------------------------------\\\\
\textit{so, we'll use that imaginary value then}\\\\
\sqrt{-5}=x\implies -5=x^2\implies 0=x^2+5\implies \boxed{y=x^2+5}[/tex]
when you get a "solution" or zero with an "i" or an imaginary value, is just a way to say, there's really no solution, the function never touches the x-axis
