найдите сумму корней уравнения
(x+3)²=16

[tex] x^{2} +6x+9-16=0[/tex]
по теореме Вьета 
[tex] \left \{ {{x1=-7} \atop {x2=1}} \right. [/tex]
x1+x2=-7+1=-6

[tex](x+3) ^{2} =16\\ x^{2} +9+6x-16=0\\ x^{2} +6x-7=0\\D=36+27=64 \\ \sqrt{D} =8\\ x_{1} = \frac{-6+8}{2} =1 \\ x_{2} = \frac{-6-8}{2} =-7\\\\ x_{1} + x_{2} =1+(-7)=1-7=-6 [/tex]