Respuesta :

[tex]\bf 5n^2+6n+7=n^2-4n\implies 5n^2+6n+7-n^2+4n=0 \\\\\\ 4n^2+10n+7=0\\\\ -------------------------------\\\\ \begin{array}{llccll} &{{ 4}}x^2&{{ +10}}x&{{ +7}}\\ &\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]

if you get a positive value, that means 2 real solutions
if you get 0, means 1 real solution only
if you get a negative value, is an imaginary root solution, same as saying no solution.
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