6x+|4-x^2|=3найти сумму корней уравнения

[latex]6x+|4-x^2|=3[/latex]1) [latex] left { {{4-x^{2} geq 0} atop {6x+4-x^2=3}} ight. [/latex][latex] left { {{-2 leq x leq 2} atop {x^2-6x-1=0, D=40}} ight. [/latex][latex] left { {{-2 leq x leq 2} atop {x_{1}= frac{6- sqrt{40}}{2}, x_{2}= frac{6+sqrt{40}}{2}}} ight. [/latex][latex] left { {{-2 leq x leq 2} atop {x_{1}=3- sqrt{10}, x_{2}=3+sqrt{10}} ight. [/latex][latex]x= 3-sqrt{10}[/latex] - корень2) [latex]left { {{4-x^{2}leq0} atop {6x-4+x^2=3}} ight.[/latex][latex]left { {{x leq -2, x geq 2} atop {x^2+6x-7=0, D=64}} ight.[/latex][latex]left { {{x leq -2, x geq 2} atop {x_{3}=-7, x_{4}=1}} ight.[/latex][latex]x=-7[/latex] - кореньСумма корней: [latex]3-sqrt{10}-7=-4-sqrt{10}[/latex]