# How do you find the discriminant and how many solutions does n^2-3n-40=0 have?

May 3, 2018

Given: ${n}^{2} - 3 n - 40 = 0$

$\textcolor{b l u e}{\text{General observation (reading the equation)}}$

As the ${n}^{2}$ term is positive the graph is of generic shape $\cup$

The $n$ term is negative so the the vertex is to the right of the y-axis thus ${n}_{\text{vertex}} > 0$

As the vertex is to the right of the y-intercept then the ${y}_{\text{vertex") < y_("intercept}}$ and the graph 'crosses the n-axis.

Thus there are two solutions to ${n}^{2} - 3 n - 40 = 0$
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$\textcolor{b l u e}{\text{Deriving the determinant and thus the count of solutions}}$

Consider the standard form $0 = y = a {x}^{2} + b x + c$

where $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The determinant part is ${b}^{2} - 4 a c$

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

${b}^{2} - 4 a c \to {\left(- 3\right)}^{2} - 4 \left(1\right) \left(- 40\right) = 169$

As the determinant is greater than 0 it also tells us that
$\textcolor{b r o w n}{\underline{\text{there are 2 solutions.}}}$