\(x^2+4x+1=0\)
\(\Leftrightarrow x^2+4x+4-3=0\)
\(\Leftrightarrow\left(x+2\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{3}\\x+2=-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}-2\\x=-\sqrt{3}-2\end{matrix}\right.\)
Vậy \(x_1=\sqrt{3}-2;x_2=-\sqrt{3}-2\).
\(\Delta=4^2-4\cdot1\cdot1=16-4=12\)
\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{\sqrt{12}-4}{2}=\dfrac{2\left(\sqrt{3}-2\right)}{2}=\sqrt{3}-2\\ x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\sqrt{12}-4}{2}=\dfrac{2\left(-\sqrt{3}-2\right)}{2}=-\sqrt{3}-2\)
