Given: #n^2-3n-40=0#

#color(blue)("General observation (reading the equation)")#

As the #n^2# term is positive the graph is of generic shape #uu#

The #n# term is negative so the the vertex is to the right of the y-axis thus #n_("vertex")>0#

As the vertex is to the right of the y-intercept then the #y_("vertex") < y_("intercept")# and the graph 'crosses the n-axis.

Thus there are two solutions to #n^2-3n-40=0#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Deriving the determinant and thus the count of solutions")#

Consider the standard form #0=y=ax^2+bx+c#

where #x=(-b+-sqrt(b^2-4ac))/(2a)#

The determinant part is #b^2-4ac#

In this case: #a=+1; b=-3 and c=-40# giving:

#b^2-4ac ->(-3)^2-4(1)(-40) = 169#

As the determinant is greater than 0 it also tells us that

#color(brown)(ul("there are 2 solutions."))#