(х+5)^2=2 решите уравнение

[latex](x+5)^2=2\\ x+5=- \sqrt{2} \ \ \ x+5= \sqrt{2} \\ x=- \sqrt{2}-5 \ \ \ x= \sqrt{2} -5[/latex]

[latex](x+5)^{2} =2[/latex] [latex] x^{2} +10x+25=2[/latex] [latex] x^{2} +10x+25-2=0[/latex] [latex] x^{2} +10x+23=0[/latex] [latex]D=b^{2} - 4ac = 10^{2} - 4*1*23=100-92= 8 [/latex] [latex] \sqrt{D} = \sqrt{8}[/latex] [latex] x_{1,2}= \frac{-10+- \sqrt{8} }{2} = \frac{10+- \sqrt{2} }{2} [/latex] [latex] x_{1}= -5+ \sqrt{2} [/latex] [latex] x_{2} = =-5- \sqrt{2} [/latex] Ответ:x∈(-бесконечность ; - 5 - [latex] \sqrt{2} [/latex]) U (-5+[latex] \sqrt{2} [/latex] ; + бесконечность )  .