Recall that:
[tex]\begin{gathered} For\text{ all }l\in\mathbb{R} \\ l^2\ge0. \end{gathered}[/tex]
Notice that:
[tex]x^2-4x+7=x^2-4x+4+3=(x-2)^2+3.[/tex]
Since x-2 is a real number, then:
[tex](x-2)^2\ge0.[/tex]
Therefore:
[tex](x-2)^2+3\geq3>0.[/tex]
Therefore the given equation has no real solutions.
Answer: The given equation has no real solutions.
