Answer
Two distinct real solutions.
Work Step by Step
For $ax^{2}+bx+c=0$, we can find solutions using the
Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
The radicand in the formula is called the discriminant.
$D=b^{2}-4ac$
D is positive $\Rightarrow$ there are two distinct real solutions.
D is zero $\Rightarrow$ there is one (double) real solution.
D is negative $\Rightarrow$ the solutions are complex conjugates (not real).
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Here,
$ x^{2}-4x-5=0\quad\rightarrow \left\{\begin{array}{l}
a=3\\
b=-6\\
c=1
\end{array}\right.$
$D=b^{2}-4ac=(-4)^{2}-4(1)(-5)=16+20=36$
$D=b^{2}-4ac$ is positive $\Rightarrow$ two distinct real solutions.
