Answer
\begin{align*}
\text{Discriminant: }&
36
\\\text{Number of Real Solutions: }&
2
\end{align*}
Work Step by Step
In the given equation,
\begin{align*}
x^2-4x-5=0
,\end{align*} $a=
1
,$ $b=
-4
,$ and $c=
-5
.$ Using the Discriminant Formula which is given by $b^2-4ac,$ the value of the discriminant is
\begin{array}{l}\require{cancel}
&
(-4)^2-4(1)(-5)
\\&=
16+20
\\&=
36
\end{array}
Since the discriminant is greater than $0,$ then there are $2$ real solutions. Hence,
\begin{align*}
\text{Discriminant: }&
36
\\\text{Number of Real Solutions: }&
2
\end{align*}
