Answer
two real solutions
Work Step by Step
The given quadratic equation, $
x^2-5=0
,$ has the following coefficients:
\begin{array}{l}\require{cancel}
a=
1
\\b=
0
\\c=
-5
.\end{array}
Substituting these values into $b^2-4ac$ (or the Discriminant), then the value of the discriminant is
\begin{array}{l}\require{cancel}
(0)^2-4(1)(-5)
\\\\=
0+20
\\\\=
20
.\end{array}
Since the value of the discriminant is $\text{
greater than zero
,}$ then the given quadratic equation has $\text{
two real solutions
}$.
