Question

x la a doua minus 4x plus 4 egal cu minus 5​

Answer 1

$x^2-4x+4=-5\\ \\ x^2-4x+9=0\\ \\ \Delta=(-4)^2-4\cdot 9\cdot 1=16-36=-20\\ \\ x_{1, \ 2}=\frac{-(-4)\pm \sqrt{-20}}{2\cdot 1}=\frac{4\pm \sqrt{2^2\cdot 5\cdot i^2}}{2}=\frac{2^2\pm 2i\sqrt{5}}{2}=2\pm i\sqrt{5}\\ \\ x_1=2+\sqrt{5}\\ \\ x_2=2-\sqrt{5}\\ \\ S=\{2-\sqrt{5}, \ 2+\sqrt{5} \}$

Answer 2

$\it x^2-4x+4=-5 \Rightarrow (x-2)^2=-5 \Rightarrow\sqrt{(x-2)^2}=\sqrt{-5} \Rightarrow\\ \\ \Rightarrow |x-2|=i\sqrt5 \Rightarrow x-2 = \pm i\sqrt5 \Rightarrow x-2\in\{-i\sqrt5,\ \ i\sqrt5\}|_{+2} \Rightarrow \\ \\ \Rightarrow x\in\{2-i\sqrt5,\ \ 2+i\sqrt5\}$