$$ x^{2}+2 x+5=0 $$ many distinct real solutions does the given equation have?

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Solve the equation $$ x^{2}+2x+5=0 $$.
Solve the equation(The complex numbers system) by following steps:
- step0: Solve using the quadratic formula in the complex numbers system:
  $$x^{2}+2x+5=0$$
- step1: Solve using the quadratic formula:
  $$x=\frac{-2\pm \sqrt{2^{2}-4	imes 5}}{2}$$
- step2: Simplify the expression:
  $$x=\frac{-2\pm \sqrt{-16}}{2}$$
- step3: Simplify the expression:
  $$x=\frac{-2\pm 4i}{2}$$
- step4: Separate into possible cases:
  $$\begin{align}&x=\frac{-2+4i}{2}\&x=\frac{-2-4i}{2}\end{align}$$
- step5: Simplify the expression:
  $$\begin{align}&x=-1+2i\&x=\frac{-2-4i}{2}\end{align}$$
- step6: Simplify the expression:
  $$\begin{align}&x=-1+2i\&x=-1-2i\end{align}$$
- step7: Rewrite:
  $$x_{1}=-1-2i,x_{2}=-1+2i$$
- step8: Remove the complex number(s):
  $$	extrm{No real solution}$$
The given equation $$x^{2}+2x+5=0$$ has no real solutions.
The equation $$ x^{2}+2x+5=0 $$ has no real solutions.

many distinct real solutions does the given equation have?

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