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